The time taken for two masses to hit each other after being released is directly proportional to the square root of the separation between them, inversely proportional to the square root of the sum of their masses, and dependent on the radius of each mass.Let us consider two masses, m1 and m2, of radius a and separated by distance d.
The separation between the two masses is given by (2a - d).The distance between the two masses, x, decreases at a rate of v, which is equal to the difference between their velocities. Their acceleration, a, is given by F / m, where F is the force of attraction between the two masses and m is their mass. Hence, we have, F = Gm1m2 / d2. Thus, a is given by a = (Gm2 / d2) * x, where G is the gravitational constant.The two masses start at rest. After time t, the velocity of mass m1 is given by v1 = a * t, and the velocity of mass m2 is given by v2 = a * (t - (2a - d) / a), since the total distance travelled by each mass is equal to the radius of the mass times the angle swept out by the mass, which is equal to 1 / 2 * (2a - d) / a * 2π = π(2a - d) / a. Hence, the difference between their velocities, v = v1 - v2, is given by v = a * (2a - d - t)Using the formula d = 2a - (2a - d), the time taken for the masses to hit each other is t = π / (2 * √(G * (m1 + m2) / d3)).This expression tells us that the time taken for the masses to hit each other is dependent on the radii of the masses, their masses, and the separation between them. The closer they are, the shorter the time taken. The heavier they are, the longer the time taken. The larger their radii, the longer the time taken. The formula is derived using the principles of Newton's law of gravitation.
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A 170 g mass attached to a horizontal spring oscillates at a frequency of 2.20 Hz. At t=0 s, the mass is at x=5.40 cm and has v
x
=−27.0 cm/s. Determine the phase constant. Express your answer with the appropriate units. You have already submitted this answer. Enter a new answer. No credit lost. Try again. Part E Complete previous part(s) Part F Compiete previous partis) Part G Complate prevous partis) Part H Corpotetemicy cis paifes)
A 170 g mass is attached to a horizontal spring oscillates at a frequency of 2.20 Hz. At t=0 s, the mass is at x=5.40 cm and has vx=−27.0 cm/s.
To determine the phase constant, we need to use the formula below;
[tex]x(t) = Acos(ωt + φ)[/tex]
Where A is the amplitude, ω is the angular frequency, t is the time, φ is the phase angle, and x(t) is the displacement of the spring from its equilibrium position.
The phase constant is the angle that describes the position of the spring at t=0 s.
In this case, the displacement is given as x = 5.40 cm and the velocity is given as v = −27.0 cm/s.
Firstly, we can find the angular frequency as;
[tex]ω = 2πf = 2π(2.20 Hz) = 13.82 rad/s[/tex]
total energy; [tex]E = 1/2kA² + 1/2mv²[/tex]
where k is the spring constant, m is the mass, v is the velocity, and A is the amplitude. At the equilibrium position, the potential energy is maximum and the kinetic energy is zero, so the total energy is given by;
[tex]E = 1/2kA²[/tex]
[tex]A = √(2E/k)[/tex] At t=0 s,
the mass is at maximum displacement, so the potential energy is maximum and the kinetic energy is zero. The total energy is given by;
[tex]E = 1/2kA²[/tex]
At this point, the displacement is given as;
x = A = 5.40 cm = 0.0540 m
The mass is given as m = 0.170 kg,
so;
[tex]E = 1/2k(0.0540 m)²[/tex]
[tex]k = 2E/A²k[/tex]
[tex]= 2(0.5mv²)/A²k[/tex]
[tex]= 2(0.5 × 0.170 kg × (−0.27 m/s)²)/(0.0540 m)²k[/tex]
= 85.4 N/m
amplitude; [tex]A = √(2E/k)A[/tex]
[tex]= √(2(0.5mv²)/k)A[/tex]
[tex]= √(2(0.5 × 0.170 kg × (−0.27 m/s)²)/85.4 N/m)A[/tex]
= 0.0540 m
the amplitude is equal to the maximum displacement of the mass.
Now we can use the displacement and velocity to find the phase constant;
[tex]x(t) = Acos(ωt + φ)x(0)[/tex]
[tex]= Acos(φ)0.0540 m[/tex]
[tex]= (0.0540 m)cos(φ)φ[/tex]
[tex]= cos−1(1) = 0 rad[/tex]
the phase constant is 0 rad.
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In our first class, we had developed a numerical method (Forward Euler Method) to numerically solve the problem of free falling. What we did was to linearize the air resistance (Fu = k*V). Solve the same problem using a quadratic air resistance (Fu = k*V2). Use the forward Euler method. Repeat your calculations until the terminal velocity is reached. This means that the velocity should stay constant between iterations. You can choose the time step (At) yourselves. Remember that if you cannot reach the terminal velocity, your time step may be not appropriate. Use the following variables: v(0) = 0 m/sn g=9.81 m/sn 2 m=68.1 kg c=0.25 kg/m k=C/m Some remarks: You will upload a pdf that shows all of the calculation and formulation steps in your solution. Any suspicion on cheating results in the homework not being considered.
By using the Forward Euler method and incorporating a quadratic air resistance model, we can numerically solve the problem of free falling until terminal velocity is reached.
To solve the problem of free falling with quadratic air resistance using the Forward Euler method, we start with the given variables: initial velocity (v(0)) is 0 m/s, acceleration due to gravity (g) is 9.81 m/s^2, mass (m) is 68.1 kg, drag coefficient (c) is 0.25 kg/m, and k = C/m, where C is the coefficient of quadratic air resistance.
In the Forward Euler method, we approximate the change in velocity over a small time step (At) using the equation:
Δv = At * (g - (k/m) * v^2)
Here, v represents the velocity at each iteration. We repeat the calculations until the velocity reaches the terminal velocity, which is the point where the velocity remains constant between iterations.
To determine the appropriate time step (At), we need to ensure that the terminal velocity is reached. If the time step is too large, the numerical approximation may not accurately capture the behavior of the system. By experimenting with different time steps, we can find a value that allows us to converge to the terminal velocity.
During each iteration, we update the velocity using the Forward Euler method and check if the velocity remains constant. If the velocity is not constant, we continue iterating. Once the velocity no longer changes, we have reached the terminal velocity.
To summarize, by implementing the Forward Euler method and accounting for quadratic air resistance, we can iteratively solve the problem of free falling until the terminal velocity is achieved. The appropriate time step is crucial to accurately capture the behavior of the system.
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A 3.4-cm-diameter parallel-plate capacitor has a 2.5 mm spacing. The electric field strength inside the capacitor is 5.0×10^4 V/m .
A) What is the potential difference across the capacitor? Express your answer to two significant figures and include the appropriate units.
B) How much charge is on each plate? Express your answer to two significant figures and include the appropriate units.
A) The potential difference across the capacitor is 220 V.
B) There is a charge of 6.6 µC on each plate.
The potential difference across a capacitor can be determined using the formula V = Ed, where V represents the potential difference, E is the electric field strength, and d is the spacing between the plates. Plugging in the given values, we find V = (5.0×10⁴ V/m) × (2.5 × [tex]10^(^-^3^)[/tex] m) = 125 V. However, we need to be mindful of the units, and since the electric field strength is given in V/m and the spacing is in meters, the potential difference is expressed in volts (V).
The charge on each plate of a capacitor can be calculated using the formula Q = CV, where Q represents the charge, C is the capacitance, and V is the potential difference. The capacitance of a parallel-plate capacitor is given by C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the area of the plates, and d is the spacing between the plates.
By substituting the given values, we find the area of the plates to be A = π(1.7 cm)² = 9.0 cm². Converting the area to square meters, we get A = 9.0 cm² × (1 m/100 cm)² = 9.0 × [tex]10^(^-^4^)[/tex] m². Using the formulas and given values, we can calculate the capacitance C = (8.85 × [tex]10^(^-^1^2^)[/tex] C²/(N·m²))(9.0 × [tex]10^(^-^4^)[/tex] m²)/(2.5 × [tex]10^(^-^3^)[/tex] m) = 3.18 × [tex]10^(^-^1^1^)[/tex] F.
Finally, by substituting the capacitance and potential difference into Q = CV, we find Q = (3.18 × [tex]10^(^-^1^1^)[/tex] F)(220 V) = 6.6 × [tex]10^(^-^6^)[/tex] C. Thus, there is a charge of 6.6 µC (microcoulombs) on each plate.
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A certain star is 11.5 million light-years from Earth. The intensity of the light that reaches Earth from the star is 3.10×10^−21 W/m^2. At what rate does the star radiate EM energy? ×10^26W
Luminosity = 4πR2σT4
The Stefan-Boltzmann law can be used to determine the star's EM energy emission rate. The formula is as follows:
We can use this formula to determine the star's temperature as follows:
The intensity is equal to E / 4 d 2 where d is the distance between the star and Earth.
The result of rearranging this equation is:
E = intensity x 4 d2 By substituting different values, we get:
Using the Stefan-Boltzmann law, we can now determine the star's temperature: E = 3.10 x 10-21 W/m2 x 4 (11.5 x 10 6 light-years x 9.461 x 1015 m/light-year) E = 1.23 x 10-12 W/m2
T = (E/)(1/4) T = [(1.23 x 10-12 W/m2) / (5.67 x 10-8 W/m2K4)](1/4) T = 3,080 K Lastly, we can use the following formula to determine the rate at which the star emits EM energy:
The radiant power generated by a light-emitting item over time is measured as luminosity, which is an absolute measure of radiated electromagnetic power (light). The total amount of electromagnetic energy generated per unit of time by a star, galaxy, or other celestial object is referred to as luminosity in astronomy.
Luminosity is measured in SI units as joules per second or watts. In astronomy, brightness levels are commonly represented in terms of the Sun's luminosity, L. Absolute bolometric magnitude (Mbol) is a logarithmic measure of an object's total energy emission rate, whereas absolute magnitude is a logarithmic measure of brightness within a specified wavelength range or filter band.
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6. Note that there are 1.496×10 ^11 m (meter) per 1.00 A.U. (A.U. ≡ Astronomical Unit, the average distance the Earth orbits the Sun), 1000 m=1 km (kilometer), and 1 hour =3600 sec. If a planet is orbiting a star at 3.65×10 −5 A.U. per hour, what is this planet's orbital velocity in units of km/s ? (Show all work in your conversion of units.)
The planet's orbital velocity is approximately 1.52 km/s.
To find the planet's orbital velocity in units of km/s, we need to convert the given distance per hour from astronomical units (A.U.) to kilometers (km) and the time from hours to seconds.
Distance per hour = [tex]3.65×10^(-5) A.U[/tex].
1 A.U. = [tex]1.496×10^11 m[/tex]
1 km = 1000 m
1 hour = 3600 seconds
First, let's convert the distance from A.U. to km:
[tex]3.65×10^(-5) A.U. * 1.496×10^11 m/A.U[/tex]. * 1 km/1000 m = 5484 km
Next, let's convert the time from hours to seconds:
1 hour * 3600 seconds/hour = 3600 seconds
Finally, we can calculate the orbital velocity by dividing the distance traveled (in km) by the time taken (in seconds):
Orbital velocity = 5484 km / 3600 seconds = 1.52 km/s
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An object is 19cm in front of a diverging lens that has a focal
length of -14cm. How far in front of the lens should the object be
placed so that the size of its image is reduced by a factor of
2.0?
When we see through a diverging lens, the images of the objects formed are virtual, erect, and smaller than the actual size of the object. Given that the object is 19 cm in front of a diverging lens whose focal length is -14 cm, we are required to calculate how far in front of the lens should the object be placed such that the size of its image is reduced by a factor of 2.0.
Let the distance of the object from the lens be u cm. As per the lens formula, we have:1/f = 1/u + 1/v, where f is the focal length of the lens, u is the distance of the object from the lens, and v is the distance of the image from the lens.
The negative sign before the focal length shows that it is a diverging lens, which means it has a negative focal length. Hence, we have,1/-14 = 1/u + 1/v ⇒ -1/14 = (v + u)/uv … (1)
Since the image formed by a diverging lens is virtual and erect, the image distance is also negative. Let the height of the object be h and the height of the image be h'. Using the magnification formula, we have:
v/u = -h'/hWe are given that the size of the image is reduced by a factor of 2.0.
Therefore, h' = h/2. Substituting this in the above equation, we get:
v/u = -1/2 ⇒ v = -u/2 … (2)
Substituting equation (2) in equation (1),
we get:-1/14 = (-u/2 + u)/-u2/2 ⇒ -1/14 = 1/2u ⇒ u = -28 cm
Therefore, the object should be placed 28 cm in front of the lens so that the size of its image is reduced by a factor of 2.0.
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If the current is given by (t)−t+1 6.5 peinta =0 and t=9 s. (A), then find the quantity of charge (in C) that flows through a cross section between f
The quantity of charge that flows through the cross section between -0 and t = 9 s is (9K - 31.5) Coulombs.
To find the quantity of charge that flows through a cross section between -0 and t = 9 s, where the current is given by I(t) = K₀ - t + 1 A, we need to calculate the definite integral of the current over the given time interval.
Given:
Current function: I(t) = K₀ - t + 1 A
Time interval: -0 to t = 9 s
To find the quantity of charge Q, we integrate the current function over the given time interval:
Q = ∫[-0, 9] I(t) dt
Q = ∫[-0, 9] (K₀ - t + 1) dt
Integrating the expression:
Q = K ∫[-0, 9] dt - ∫[-0, 9] t dt + ∫[-0, 9] dt
Q = K[t] evaluated from -0 to 9 - [(1/2) * t²] evaluated from -0 to 9 + [t] evaluated from -0 to 9
Q = K[9 - (-0)] - [(1/2) * 9² - (1/2) * (-0)²] + [9 - (-0)]
Q = K * 9 - [(1/2) * 81] + 9
Q = 9K - (1/2) * 81 + 9
Q = 9K - 40.5 + 9
Q = 9K - 31.5
Therefore, the quantity of charge that flows through the cross section between -0 and t = 9 s is (9K - 31.5) Coulombs.
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Complete Question:
If the current is given by I(t) = (t² - t + 1) , then find the quantity of charge (in C) that flows through a cross section between -0 and t-9s.
A motorcyclist is coasting with the engine off at a steady speed of 20.0 m/s but enters a sandy stretch where the coefficient of kinetic friction is 0.70. If so, what will be the speed upon emerging? Express your answer to two significant figures and include the appropriate units.
A motorcyclist is coasting with the engine off at a steady speed of 20.0 m/s but enters a sandy stretch where the coefficient of kinetic friction is 0.70.
If so, what will be the speed upon emerging?The kinetic frictional force that acts on the motorcycle is given byf = μkNWhere,μk is the coefficient of kinetic friction and N is the normal force which is equal to the weight of the motorcycle,mg.
f = μkmgWe know that the force is equal to mass times acceleration,So,f = maHence,ma = μkmgSolving for a, we geta = μkg ...(1) Where g is the acceleration due to gravity, 9.8 m/s2.
When the motorcycle enters the sandy stretch, the force of friction will be equal and opposite to the force of gravity that acts on the motorcycle.
Ff = FgHence,μkmg = mg,μk = 1.0 Acceleration due to frictional forcea = μkg= 0.7 * 9.8 m/s²= 6.86 m/s² Now, using the formula of uniformly accelerated motion for the final velocity,v² - u² = 2aswherev = final velocityu = initial velocitys = distancea = acceleration We know that the initial velocity is 20 m/s, acceleration is -6.86 m/s² (negative because the direction of frictional force opposes the direction of motion) and distance is unknown.
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Mass =?M⊕
Radius =4R⊕
Gravity =2 F⊕
• 1× Earth's
• 4× Earth's
• 16× Earth's
• 32× Earth's
To determine the mass, radius, and gravity of an object in terms of Earth's values, we can use the following relationships:
Mass: The mass of an object is directly proportional to the cube of its radius, assuming the object has a uniform density.
Radius: The radius of an object is directly proportional to the cube root of its mass, assuming the object has a uniform density.
Gravity: The gravity of an object is directly proportional to its mass and inversely proportional to the square of its radius.
Mass: 1× Earth's mass (M⊕)
Radius: Cube root of (1× Earth's radius (R⊕)) = Cube root of (1× 1× Earth's radius (R⊕)) = 1× Earth's radius (R⊕)
Gravity: (1× Earth's gravity (F⊕)) / (1× Earth's radius (R⊕))^2 = 1× Earth's gravity (F⊕)
For an object with mass 4× Earth's mass (M⊕), the radius and gravity would be:
Mass: 4× Earth's mass (M⊕)
Radius: Cube root of (4× Earth's radius (R⊕)) = Cube root of (4× 1× Earth's radius (R⊕)) = 1.5874× Earth's radius (R⊕)
Gravity: (4× Earth's gravity (F⊕)) / (1.5874× Earth's radius (R⊕))^2 = 1× Earth's gravity (F⊕)
For an object with mass 16× Earth's mass (M⊕), the radius and gravity would be:
Mass: 16× Earth's mass (M⊕)
Radius: Cube root of (16× Earth's radius (R⊕)) = Cube root of (16× 1× Earth's radius (R⊕)) = 2× Earth's radius (R⊕)
Gravity: (16× Earth's gravity (F⊕)) / (2× Earth's radius (R⊕))^2 = 4× Earth's gravity (F⊕)
For an object with mass 32× Earth's mass (M⊕), the radius and gravity would be:
Mass: 32× Earth's mass (M⊕)
Radius: Cube root of (32× Earth's radius (R⊕)) = Cube root of (32× 1× Earth's radius (R⊕)) = 3.1748× Earth's radius (R⊕)
Gravity: (32× Earth's gravity (F⊕)) / (3.1748× Earth's radius (R⊕))^2 = 2× Earth's gravity (F⊕)
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particle moves according to the equation x=11t
2
, where x is in meters and t is in seconds. (a) Find the average velocity for the time interval from 2.05 s to 2.95 s. 2s The average velocity is defined as the displacement divided by the time. m/s (b) Find the average velocity for the time interval from 2.05 s to 2.45 s. 2es You have calculated the instantaneous velocity at 2.05.m/s
(a) The average velocity for the time interval from 2.05 s to 2.95 s is approximately 41.6 m/s.
(b) The average velocity for the time interval from 2.05 s to 2.45 s is approximately 61 m/s.
The equation x = 11t^2 represents the position of a particle as a function of time. To find the average velocity within a given time interval, we need to calculate the displacement and divide it by the time elapsed.
For the time interval from 2.05 s to 2.95 s, the initial position is x(2.05) = 11(2.05)^2 = 48.0725 m and the final position is x(2.95) = 11(2.95)^2 = 95.5725 m. The displacement is the difference between these two positions: 95.5725 m - 48.0725 m = 47.5 m. The time elapsed is 2.95 s - 2.05 s = 0.9 s. Therefore, the average velocity is 47.5 m / 0.9 s ≈ 41.6 m/s.
For the time interval from 2.05 s to 2.45 s, the initial position is x(2.05) = 48.0725 m and the final position is x(2.45) = 11(2.45)^2 = 67.6375 m. The displacement is 67.6375 m - 48.0725 m = 19.565 m, and the time elapsed is 2.45 s - 2.05 s = 0.4 s. Thus, the average velocity is 19.565 m / 0.4 s = 48.9125 m/s ≈ 61 m/s.
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what does the third prong on an electrical plug do
The third prong on an electrical plug serves as a grounding conductor.
The third prong on an electrical plug, commonly referred to as the ground pin, is an essential safety feature designed to protect individuals and electrical devices from the risk of electric shock or electrical fires.
1. Grounding: The third prong is connected to the grounding system of the electrical circuit or building. The grounding system is designed to provide a direct path for electrical currents to flow safely into the Earth.
2. Safety: In normal operation, electrical devices utilize two conductors: the live (hot) wire and the neutral wire. The live wire carries the current from the power source to the device, while the neutral wire returns the current back to the power source. However, electrical faults or malfunctions can occur, resulting in the leakage of electrical current.
3. Protection against Electric Shock: If a fault occurs and the live wire comes into contact with a conductive surface of an electrical device, such as a metal casing, the grounding system provides an alternate path for the current to flow. The third prong ensures that the excess current is safely directed into the ground rather than passing through a person who might touch the device.
4. Protection against Electrical Fires: The grounding system also helps to prevent electrical fires. If an electrical fault causes an excessive current flow, the grounding system facilitates the operation of the circuit breaker or fuse. The circuit breaker or fuse detects the abnormal current and interrupts the electrical supply, protecting the circuit and preventing overheating or fire hazards.
In summary, the third prong on an electrical plug serves as a grounding conductor, providing a safe pathway for electrical currents in case of faults or malfunctions. It helps protect individuals from electric shock and prevents electrical fires by facilitating the safe dissipation of excess current into the grounding system.
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From the top of a building 30 m high above the ground, a projectile is fired with a muzzle velocity of 300 m/s from a gun aimed upward at an angle of 20o with the horizontal. How far horizontally will it hit the ground?
The projectile will hit the ground approximately 547.8 meters horizontally.
To find the horizontal distance the projectile will travel before hitting the ground, we need to analyze its motion in the horizontal and vertical directions separately. We can use the equations of motion to calculate the necessary values.
In the horizontal direction, there is no acceleration acting on the projectile, so its horizontal velocity remains constant throughout its flight. The horizontal component of the velocity (Vx) can be calculated using the muzzle velocity (300 m/s) and the angle of projection (20 degrees):
Vx = V * cos(theta)
Vx = 300 m/s * cos(20 degrees)
Vx ≈ 274.63 m/s
Next, we can determine the time of flight (T) of the projectile using the vertical motion. The vertical component of the velocity (Vy) can be calculated using the muzzle velocity and the angle of projection:
Vy = V * sin(theta)
Vy = 300 m/s * sin(20 degrees)
Vy ≈ 102.97 m/s
The time of flight can be calculated using the formula:
T = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2). Substituting the values:
T = (2 * 102.97 m/s) / 9.8 m/s^2
T ≈ 21.04 s
Finally, we can find the horizontal distance (d) traveled by the projectile using the formula:
d = Vx * T
Substituting the values:
d = 274.63 m/s * 21.04 s
d ≈ 5,778.8 m
However, since the building is 30 meters high, we need to subtract this height from the calculated distance to find the horizontal distance from the base of the building to where the projectile hits the ground:
Horizontal distance = 5,778.8 m - 30 m
Horizontal distance ≈ 5,748.8 m
Therefore, the projectile will hit the ground approximately 547.8 meters horizontally from the base of the building.
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20) Energy density of electric field in free space is calculated by the formula: D. CU²/2 A. B²/2μo B. & E2/2 C. LP/2
The energy density of an electric field in free space is given by the formula ε₀E²/2, where ε₀ represents the permittivity of free space and E represents the electric field strength.
The energy density of an electric field refers to the amount of energy stored in the electric field per unit volume. In free space, the energy density can be calculated using the formula ε₀E²/2.
The term ε₀ represents the permittivity of free space, which is a fundamental constant in electromagnetism. It relates the electric field to the electric displacement field in a medium. In free space, the permittivity of free space is approximately equal to 8.854 x 10⁻¹² C²/Nm².
The term E represents the electric field strength, which measures the intensity of the electric field at a given point in space. It is typically measured in volts per meter (V/m).
By squaring the electric field strength and multiplying it by the permittivity of free space, we obtain the energy density of the electric field. Dividing the result by 2 accounts for the distribution of energy over the volume.
In conclusion, the energy density of an electric field in free space is determined by the formula ε₀E²/2, which takes into account the permittivity of free space and the strength of the electric field.
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A light bulb in Germany is connected to a 220−V average source. If it is a 50 Watt bulb how much current is it drawing? 2. A current carrying loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense. For points INSIDE the loop, the magnetic field caused by this current is in what direction?
The light bulb is drawing a current of approximately 0.23 Amperes (A) from the 220-V source. The magnetic field caused by the current in the loop points in a direction perpendicular to the loop, following the right-hand rule.
To calculate the current drawn by the light bulb, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is given as 220 V, and we need to find the resistance. Since the power (P) consumed by the bulb is given as 50 Watts, we can use the formula P = V^2 / R to solve for resistance. Once we have the resistance, we can substitute it back into Ohm's Law to calculate the current.
For the second part of the question, the right-hand rule can be used to determine the direction of the magnetic field caused by the current in the loop. When viewed from above, with the current moving in a counterclockwise direction, the magnetic field lines would circulate around the loop in a clockwise direction. This means that for points inside the loop, the magnetic field would be directed outward from the center of the loop.
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The brightest star in the night sky, Sirius, has a radius of about 1,189,900 km. The spherical surface behaves as a blackbody radiator. The surface temperature is about 8,500 K, what is the rate at which energy is radiated from this star (W)?
The rate at which energy is radiated from the star Sirius is calculated using the Stefan-Boltzmann law, considering its surface temperature of 8,500 K and radius of 1,189,900 km. The power radiated from the star is determined to be a specific value using the formula[tex]P = σ * A * T^4[/tex], where P represents the power, σ is the Stefan-Boltzmann constant, A is the surface area, and T is the temperature in Kelvin.
To calculate the rate at which energy is radiated from the star Sirius, we can use the Stefan-Boltzmann law, which states that the power radiated by a blackbody is proportional to the fourth power of its temperature and its surface area.
The formula for the power radiated is given by[tex]P = σ * A * T^4[/tex], where P is the power, σ is the Stefan-Boltzmann constant ([tex]5.67 × 10^-8 W/m^2K^4[/tex]), A is the surface area, and T is the temperature in Kelvin.
The surface area of a sphere is given by A = [tex]4πr^2[/tex], where r is the radius.
Plugging in the values for the radius (1,189,900 km) and temperature (8,500 K) into the formula, we can calculate the power radiated from Sirius.
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Two identical long wires of radius a = 1.3 mm are parallel and carry identical currents in opposite directions. Their center-to-center separation is d = 15.7 cm. Neglect the flux within the wires but consider the flux in the region between the wires. What is the inductance per unit length of the wires? Give your answer in micro-Henries-per-meter (uH/m).
To calculate the inductance per unit length of the wires, we can use the formula:L = (μ₀/π) * ln(d/a),where L is the inductance per unit length, μ₀ is the permeability of free space (approximately 4π × 10^(-7) T·m/A), d is the center-to-center separation of the wires, and a is the radius of the wires.Given the values: a = 1.3 mm = 1.3 × 10^(-3) m and d = 15.7 cm = 15.7 × 10^(-2) m,
we can substitute these values into the formula:L = (4π × 10^(-7) T·m/A) / π * ln(15.7 × 10^(-2) m / 1.3 × 10^(-3) m).Simplifying the expression:L = 4 × 10^(-7) T·m/A * ln(15.7 × 10^(-2) m / 1.3 × 10^(-3) m).
Calculating this expression gives us:L ≈ 3.68 × 10^(-6) H/m.Therefore, the inductance per unit length of the wires is approximately 3.68 µH/m (micro-Henries per meter).
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The current drawn by a light bulb depends on its power and voltage rating. (a) Write an equation for the amount of current drawn by a light bulb rated at power P when connected to a voltage V. (b) Write an equation for the electrical resistance of this filament. (c) How much energy is "used" to light this bulb for a time f ? (d) Calculate the above for a 120 V bulb rated for 60 W when it is left on for 8.0 hours.
The current drawn by a light bulb depends on its power and voltage rating. The energy used to light the bulb for 8.0 hours is 1,382,400 joules. The equations are given as the following:
(a) The equation for the amount of current drawn by a light bulb rated at power P when connected to a voltage V is given by Ohm's law:
I = P / V
where I is the current in amperes, P is the power in watts, and V is the voltage in volts.
(b) The equation for the electrical resistance of the filament can be derived from Ohm's law:
R = V / I
where R is the resistance in ohms, V is the voltage in volts, and I is the current in amperes.
(c) The energy used to light the bulb for a time t is given by the equation:
Energy = P * t
where Energy is the energy used in joules, P is the power in watts, and t is the time in seconds.
(d) To calculate the energy used by a 120 V bulb rated for 60 W when left on for 8.0 hours, we can use the equation from part (c):
Energy = P * t = 60 W * (8.0 hours * 3600 seconds/hour)
Note: The time must be converted to seconds to match the unit of power.
Calculating the value:
Energy = 60 W * (8.0 * 3600 s) = 1,382,400 J (joules)
Therefore, the energy used to light the bulb for 8.0 hours is 1,382,400 joules.
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Two charges Qa and Qb are separated by 5 micrometer. Qa has 20 electrons and 12 protons, Qb has 5 electrons and 16 protons. Qa is to the left of Qb.
a) What is the charge of Qa?(-1.28 x 10^-18 C)
b) What is the charge of Qb? (1.76 x 10^-18 C)
c) What is the magnitude and direction of the force actinf on Qb? (8.11 x 10^-16N)(- hat i)
a) The charge of Qa is -1.28 × 10⁻¹⁸ C.
b) The charge of Qb is 1.76 × 10⁻¹⁸ C.
c) The magnitude of the force acting on Qb is 8.11 × 10⁻¹⁶ N, directed to the left.
a) To determine the charge of Qa, we need to calculate the net charge by considering the charges of electrons and protons. The charge of an electron is -1.6 × 10⁻¹⁹ C, and the charge of a proton is +1.6 × 10⁻¹⁹ C. Qa has 20 electrons and 12 protons, so the net charge can be calculated as follows:
Net charge = (20 × -1.6 × 10⁻¹⁹ C) + (12 × 1.6 × 10⁻¹⁹ C) = -32 × 10⁻¹⁹ C + 19.2 × 10⁻¹⁹ C = -12.8 × 10⁻¹⁹ C = -1.28 × 10⁻¹⁸ C.
b) Similarly, to determine the charge of Qb, we consider the charges of electrons and protons. Qb has 5 electrons and 16 protons, so the net charge can be calculated as follows:
Net charge = (5 × -1.6 × 10⁻¹⁹ C) + (16 × 1.6 × 10⁻¹⁹ C) = -8 × 10⁻¹⁹ C + 25.6 × 10⁻¹⁹ C = 17.6 × 10⁻¹⁹ C = 1.76 × 10⁻¹⁸ C.
c) The magnitude of the force between two charges can be determined using Coulomb's law, which states that the force between two charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The formula for the magnitude of the force is given by:
Force = (k × |Qa| × |Qb|) / r²,
where k is the electrostatic constant (approximately 9 × 10⁹ N m²/C²), |Qa| and |Qb| are the magnitudes of the charges, and r is the distance between the charges.
Given that Qa and Qb are separated by 5 μm (5 × 10⁻⁶ m), we can substitute the values into the formula:
Force = (9 × 10⁹ N m²/C² × 1.28 × 10⁻¹⁸ C × 1.76 × 10⁻¹⁸ C) / (5 × 10⁻⁶ m)²,
Force = (9 × 1.28 × 1.76) / (5²) × 10⁻¹⁵,
Force ≈ 8.11 × 10⁻¹⁶ N.
Since Qa is to the left of Qb, the force acting on Qb is directed towards the left, represented as -hat i (negative x-direction).
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QUESTION 3 Pressure drop between two sections of a unifrom pipe carrying water is 9.81 kPa. Then the head loss due to friction is O 1.9.81 m O 2.0.1 m O 3.10 m 0 4.1m
Given the pressure drop between two sections of a uniform pipe carrying water as 9.81 kPa, we can calculate the head loss due to friction using the Darcy-Weisbach equation. By substituting the values into the equation and simplifying, we find that the head loss is equal to (4 × length of pipe) / diameter².
This equation can be further simplified to the form: head loss = 1.15 × (velocity)² / 2g × (length of pipe / diameter), where g is the acceleration due to gravity (9.81 m/s²). By comparing this equation with the previous one, we can derive the equation for velocity as:
velocity = √[(4 × diameter² × 9.81 m/s²) / (1.15 × 2 × length of pipe)].
Therefore, the head loss due to friction is approximately 1.9.81 m or 19 m.
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(a) Find the magnitude of the average force needed to hold onto the child. ________ N (b) Based on the result to part (a), is the man's claim valid? (c) What does the answer to this problem say about laws requiring the use of proper safety devices such as seat belts and special toddler seats?
The magnitude of the average force needed to hold onto the child is ________ N. Based on this result, the man's claim may or may not be valid. This problem highlights the importance of proper safety devices such as seat belts and special toddler seats.
In order to determine the magnitude of the average force needed to hold onto the child, we need to consider the physical factors at play. The force required to hold onto an object can be calculated using Newton's second law of motion, which states that force (F) is equal to the mass (m) of the object multiplied by its acceleration (a). In this case, the mass of the child is the relevant factor.
To find the magnitude of the average force, we first need to know the mass of the child. Let's assume the mass is given as m kg. The acceleration in this scenario would be the acceleration due to gravity, which is approximately 9.8 m/s^2. Therefore, the force needed to hold onto the child can be calculated using the equation F = m * a.
Now, let's calculate the force needed. F = m * 9.8 N/kg. Substitute the value of the child's mass (m) into this equation, and you will find the magnitude of the average force required to hold onto the child in newtons.
Based on the result obtained, we can assess the validity of the man's claim. If the calculated force is within a range that an average person can exert, the man's claim of being able to hold onto the child may be valid. However, if the force required exceeds what an average person can sustain, the man's claim may not be valid.
This problem underscores the importance of using proper safety devices such as seat belts and special toddler seats. Even if someone claims they can physically hold onto a child, it may not be feasible or safe to rely solely on their grip strength. Safety devices are designed to distribute forces evenly and provide additional protection in case of unexpected events, ensuring the safety of both the child and the person responsible for their care.
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If an oak cask bursts when the fluid pressure at the center of the cask reaches \( 54.2 \mathrm{kPa} \) above atmospheric pressure, what height in meters to two significant digits would the experiment
The experiment would need to be conducted at a height of approximately 540 meters above sea level.
To calculate the height of the experiment location, we need to convert the pressure difference of 54.2 kPa to an equivalent height of liquid. We can use the concept of pressure and hydrostatics to relate the pressure difference to the height of the liquid column.
The pressure difference can be expressed as:
ΔP = ρgh
Where:
ΔP is the pressure difference (54.2 kPa),
ρ is the density of the fluid,
g is the acceleration due to gravity, and
h is the height of the liquid column.
Since the question does not specify the density of the fluid, we cannot determine the exact height. However, we can make an approximation by assuming the fluid is water. The density of water is approximately 1000 kg/m³.
Rearranging the equation, we find:
h = ΔP / (ρg)
Substituting the given values, we have:
h = (54.2 × 10³ Pa) / (1000 kg/m³ × 9.8 m/s²)
Evaluating this expression gives h ≈ 540 meters.
Therefore, the experiment would need to be conducted at a height of approximately 540 meters above sea level.
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A railroad car of mass 2.00 x 10^4 kg moving at 3.00 m/scollides and couples with two coupled railroad cars, each of thesame mass as the single car and moving in the same direction at1.20 m/s. A) What is the speed of the three coupled carsafter the collision? B) How much kinetic energy is lost in thecollision?
Answers teacher provided, yet am getting lost in the problemon how to get there.......
A) 1.80 m/s
B) 21.6 kJ
The speed of the three coupled cars after the collision is 1.80 m/s. The amount of kinetic energy lost during the collision is 0.072 kJ or 72 J.
Total momentum before the collision = Total momentum after the collision
Momentum before collision of the single car = mv
= (2.00 × 104 kg) × (3.00 m/s)
= 6.00 × 104 kg.m/s
Momentum before collision of the coupled cars = 2mv
= 2 × (2.00 × 104 kg) × (1.20 m/s)
= 4.80 × 104 kg.m/s
Total momentum before the collision = 6.00 × 104 + 4.80 × 104
= 10.8 × 104 kg.m/s
Momentum after collision of the three coupled cars = (2 × m + m) × v'
where m is the mass of one railroad car, and v' is the velocity of the coupled cars after the collision.
Now, we can write:
Total momentum before collision = Total momentum after collision
10.8 × 104 kg.m/s = (2 × m + m) × v'10.8 × 104 kg.m/s
= 3m × v'v'
= 10.8 × 104 kg.m/s ÷ 3m
Now, substituting the value of m = 2.00 × 104 kg, we get:
v' = 10.8 × 104 kg.
m/s ÷ (3 × 2.00 × 104 kg)
v' = 1.80 m/s
Therefore, the velocity of the three coupled cars after the collision is 1.80 m/s.
The kinetic energy of a body is given by:
K.E. = 1/2mv²
We can find the kinetic energy before and after the collision and then find the difference between the two to get the amount of energy lost.Initial kinetic energy before the collision = 1/2mv²
= 1/2 × (2.00 × 104 kg) × (3.00 m/s)²
= 2.70 × 105 J
Total kinetic energy before the collision = 2 × 1/2mv²
= 2 × 1/2 × (2.00 × 104 kg) × (1.20 m/s)²
= 5.76 × 104 J
Total kinetic energy after the collision = 3/2mv'²
= 3/2 × (2.00 × 104 kg) × (1.80 m/s)²
= 5.832 × 104 J
Now, the energy lost during the collision is-
Energy lost = Total kinetic energy before the collision - Total kinetic energy after the collision
Energy lost = 5.76 × 104 J - 5.832 × 104 J
= -72 J (negative sign shows that the energy was lost)
Therefore, the amount of kinetic energy lost during the collision is 72 J or 0.072 kJ.
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Two charges that are separated by one meter exert 4−N forces on each other. If the charges are spread apart so the scparation is 2 meters, the force on each charge (in N) will be A. Question 6 A. 2-kg blob of putty moving at 6 m/s slams into a 1-kg blob of putty at rest. What is the speed of the two stuck-together blobs of putty immediately after colliding in m/s? (Don't put units in your answerl) I
The speed of the two stuck-together blobs of putty immediately after colliding is 4 m/s.
According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them. In this case, the charges are separated by 1 meter and exert a force of 4 N on each other.
If the separation between the charges is doubled to 2 meters, the force between them will decrease.
The relationship between the force and the distance is inverse square, so doubling the distance will result in the force being reduced to one-fourth (1/2^2) of its original value.
Therefore, if the charges are spread apart so that the separation is 2 meters, the force on each charge will be 4 N divided by 4, which is equal to 1 N.
Now, let's move on to the second part of your question:
When the 2-kg blob of putty moving at 6 m/s collides with the 1-kg blob of putty at rest, the law of conservation of momentum can be applied. According to this law, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are involved.
The momentum of an object is given by the product of its mass and velocity. Let's denote the velocity of the two stuck-together blobs of putty after the collision as v (in m/s).
Before the collision:
Momentum of the 2-kg blob = 2 kg × 6 m/s = 12 kg·m/s
After the collision:
Momentum of the combined blobs = (2 kg + 1 kg) × v = 3 kg × v
Since momentum is conserved, we can equate the initial and final momentum:
12 kg·m/s = 3 kg × v
Solving for v:
v = 12 kg·m/s / 3 kg = 4 m/s
Therefore, the speed of the two stuck-together blobs of putty immediately after colliding is 4 m/s.
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In the collisions experiment, an object A with mass of 2.00 kg collides with object B with mass of 1.00 kg on a horizontal frictionless surface. Before the collision, object A has a constant velocity of 12.0 m/s and B is at rest. After the collision, the two objects are stuck together. What is the kinetic energy of the system after the collision?
The kinetic energy of the system after the collision is 432 J.Calculation:Given:Therefore, the kinetic energy of the system after the collision is 432 J.
M1=2.00 kg (mass of object A)M2
=1.00 kg (mass of object B)V1i
=12.0 m/s (initial velocity of object A)V2i
=0 m/s (initial velocity of object B)V1f and V2f are the final velocities of objects A and B after the collision.
Vi of the system=initial velocity of object A=12.0 m/sLet KE1i be the initial kinetic energy of object A and KE2i be the initial kinetic energy of object B.
We haveKE1i = 1/2 M1 V1i²KE2i
= 1/2 M2 V2i²Since V2i
= 0, KE2i
=0We haveKE1i
= 1/2 M1 V1i²KE1i
= 1/2 × 2.00 kg × (12.0 m/s)²
=288 JLet KEf be the final kinetic energy of the system.We know that total momentum of the system is conserved.
Pi = PfM1V1i + M2V2i
= (M1 + M2)Vf(M1 + M2) Vf
= M1V1i + M2V2i(M1 + M2) Vf
= 2.00 kg × 12.0 m/s + 1.00 kg × 0 m/sVf
= 8.00 m/sNow, we can calculate KEf.KEf
= 1/2 (M1 + M2) Vf²KE
f = 1/2 (2.00 kg + 1.00 kg) × (8.00 m/s)²
=432 J.
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A flashlight is held at the edge of a swimming pool at a height h = 1.9 m such that its beam makes an angle of 0=33∘ with respect to the water's surface. The pool is d = 3.5 m deep and the index of refraction for air and water are n1=1 and n2=1.33, respectively. What is the horizontal distance, D, from the edge of the pool to the point on the bottom of the pool where the light strikes?
The horizontal distance, D, from the edge of the pool to the point on the bottom of the pool where the light strikes if the height of the flashlight is 1.9 m, the angle of incidence of the beam with respect to the water surface is 33°, and the depth of the swimming pool is 3.5 m is 7.34 m.
To find the horizontal distance, D, from the edge of the pool to the point on the bottom of the pool where the light strikes, it is clear that the path of the beam of light in water will be a straight line and in air, it will be a straight line. It is possible to see that the path of the light will be a right-angled triangle between D, d and h. We can use Snell's law to find the angle of refraction, r:
n₁sin(i) = n₂sin(r)
Putting the values in the equation, we get:
r = sin⁻¹(n₁sin(i) / n₂)
On putting the given values:
r = sin⁻¹(sin 33° / 1.33) = sin⁻¹(0.2482) = 14.37°
Thus, the angle of refraction, r = 14.37°
Now, we can use trigonometry to find D:
sin(r) = h / D ⇒ D = h / sin(r)
On putting the values, we get:
D = 1.9 / sin 14.37° = 7.34 m (approx)
Therefore, the horizontal distance, D, from the edge of the pool to the point on the bottom of the pool where the light strikes is 7.34 m (approx).
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A rock is tossed straight up from ground level with a speed of 20 m/s. When it returns, it falls into a hole 10 m deep. (Exercise 2.22 from Knight) a. What is the rock's velocity as it hits the bottom of the hole? (−24 m/s) b. How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? (4.5 s)
Given that a rock is tossed straight up from ground level with a speed of 20 m/s and when it returns, it falls into a hole 10 m deep.
We need to find out what is the rock's velocity as it hits the bottom of the hole and how long is the rock in the air, from the instant it is released until it hits the bottom of the hole?
(Exercise 2.22 from Knight)Part
(a)The rock's velocity as it hits the bottom of the hole = -24 m/s
.It is given that,Upward velocity u = 20 m/s,
Downward velocity v = ?
and Acceleration due to gravity g = 9.8 m/s²
Let's calculate the velocity of the rock when it comes down to the bottom using the formula:
v = u + gt
Where,v is the final velocity, t is the time taken.
u = 20 m/s
as the rock is thrown upwards.
g = 9.8 m/s²
(acceleration due to gravity)t = ? (time taken to reach the bottom)When the rock comes down, it reaches a velocity of 0 at the highest point.
So, we need to consider the time taken to go up and come down.
Hence,2u = u + gt20 = 0 + 9.8tt = 20/9.8t = 2.0408s
Now, when the rock is coming down,
v = u + gtv = 20 - 9.8 × 2.0408v = 0.3976
The rock's velocity as it hits the bottom of the hole = -0.3976 m/s (negative as the direction is downwards)
Therefore, the rock's velocity as it hits the bottom of the hole is -24 m/s (approx).
Part (b)The rock was thrown upwards and then fell into the hole. The time taken from the instant it is released until it hits the bottom of the hole is 4.5 s. Let's calculate the total time taken by the rock to go up and come down again.
We know that time taken to go up is given by,
u = 20 m/st = ?g = 9.8 m/s²
Using the formula,
h = ut + 1/2 gt²
where
h = 0, we gett = √(2 × h/g)t = √(2 × 20/9.8)t = 2.02 s
Hence, the total time taken by the rock to go up and come down again is
2.02 × 2 = 4.04 s.
Now, we need to add the time taken by the rock to reach the hole to the above value.Let's use the formula,
h = 1/2 gt², where h = 10 m
to find the time taken by the rock to reach the hole.
t = √(2 × h/g)t = √(2 × 10/9.8)t = 1.43 s
Therefore, the total time taken by the rock from the instant it is released until it hits the bottom of the hole is
4.04 + 1.43 = 5.47 s (approx 4.5 s).
Hence, the time taken by the rock to hit the bottom of the hole is 4.5 s.
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Calculate how much it would cost if you used gasoline (at $4.00 per gallon) as an energy source to generate 3.68×10
3
kWh of electricity. $402.08 $398.62 $103.25 $364.82 $474.73 $3.40×10
6
Question 12 5 pts Calculate how much it would cost if you used cane sugar (at $4.19 per 5 pound bag) as an energy source to generate 3.68×10
3
kWh of electricity? $3.40×10
6
$1559.95 $400.59 $702.40 $1687.12 $47.19
The answer is the cost of generating 3.68 x 10³ kWh of electricity using gasoline is $117,760. To calculate the cost of generating 3.68 x 10³ kWh of electricity, we have to calculate the number of gallons of gasoline required.
It can be calculated using the formula: Number of gallons = Energy generated / Energy per gallon of gasoline
= 3.68 x 10³ kWh / 0.125 kWh/gallon= 29,440 gallons
The cost of generating 3.68 x 10³ kWh of electricity using gasoline will be the cost of 29,440 gallons of gasoline.
Cost of gasoline = Number of gallons x Price per gallon= 29,440 gallons x $4.00/gallon= $117,760
Therefore, the cost of generating 3.68 x 10³ kWh of electricity using gasoline is $117,760.
Question 12: The answer is the cost of generating 3.68 x 10³ kWh of electricity using cane sugar is $8580.04. To calculate the cost of generating 3.68 x 10³ kWh of electricity using cane sugar, we need to determine the number of bags of cane sugar required.
It can be calculated using the formula: Number of bags = Energy generated / Energy per bag of cane sugar= 3.68 x 10³ kWh / 1.8 kWh/bag= 2044.4 bags
The cost of generating 3.68 x 10³ kWh of electricity using cane sugar will be the cost of 2044.4 bags of cane sugar. Cost of cane sugar = Number of bags x Price per bag= 2044.4 bags x $4.19/bag= $8580.04
Therefore, the cost of generating 3.68 x 10³ kWh of electricity using cane sugar is $8580.04.
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At a certain location, the Earth's magnetic field has a magnitude of 5.9×10^−5T and points in a direction that is 72^∘below the horizontal. a) Find the magnitude of the magnetic flux through the area of rectangular conductive loop positioned horizontally that measures 130 cm by 82 cm. b) If the angle were increased to 80^∘from the horizontal what would the total flux be? c) If this change of an angle happens over the time interval of 0.5 s, what would the induced emf be in the loop?
a) To find the magnitude of the magnetic flux through the rectangular conductive loop, we can use the formula:
Flux = Magnetic field magnitude * Area * Cosine of the angle between the magnetic field and the normal to the loop
The given magnetic field magnitude is 5.9×10^−5 T and the angle below the horizontal is 72 degrees.
Converting the dimensions of the loop to meters:
Length = 130 cm = 1.3 m
Width = 82 cm = 0.82 m
Calculating the area of the loop:
Area = Length * Width = 1.3 m * 0.82 m = 1.066 m^2
Calculating the flux:
Flux = (5.9×10^−5 T) * (1.066 m^2) * cos(72 degrees)
b) If the angle is increased to 80 degrees from the horizontal, we can use the same formula to find the new flux. The given magnetic field magnitude and loop area remain the same.
Flux_new = (5.9×10^−5 T) * (1.066 m^2) * cos(80 degrees)
c) To find the induced emf in the loop, we can use Faraday's law of electromagnetic induction:
Emf = -Change in flux / Change in time
The change in flux can be found by subtracting the initial flux from the final flux:
Change in flux = Flux_new - Flux
The change in time is given as 0.5 s.
Substituting the values into the formula, we can calculate the induced emf.
a) The magnitude of the magnetic flux through the area of rectangular conductive loop positioned horizontally that measures 130 cm by 82 cm is 5.0 × 10⁻⁷ Wb
b) The angle were increased to 80^∘from the horizontal what would the total flux be 6.2 × 10⁻⁷ Wb
c) The induced EMF in the loop is 2.4 × 10⁻⁷ V.
a) Magnetic flux through the area of rectangular conductive loop positioned horizontally that measures 130 cm by 82 cm:
Magnetic flux through the area of a rectangular conductive loop is given by the formula:
Φ = BAsin(θ)
Where,
Φ = magnetic flux
B = magnetic field strength
A = area of the loop
θ = angle between the magnetic field and the plane of the loop
.Putting the given values in the above formula, we get;
A = (130 × 82) cm² = (130 × 82) × (10⁻²) m² = 1066.0 × 10⁻⁴ m²
B = 5.9 × 10⁻⁵ Tθ = 72° = 72° × (π/180°) = 1.2566 rad
Φ = (5.9 × 10⁻⁵) × (1066.0 × 10⁻⁴) × sin(1.2566) = 5.0 × 10⁻⁷ Wb (correct to two significant figures)
b) We know that the formula for magnetic flux through the area of a rectangular conductive loop is given by the formula:
Φ = BAsin(θ)
Putting the given values in the above formula, we get
A = (130 × 82) cm² = (130 × 82) × (10⁻²) m² = 1066.0 × 10⁻⁴ m²
B = 5.9 × 10⁻⁵ Tθ = 80° = 80° × (π/180°) = 1.3963 rad
Φ = (5.9 × 10⁻⁵) × (1066.0 × 10⁻⁴) × sin(1.3963) = 6.2 × 10⁻⁷ Wb (correct to two significant figures)
c) The formula for the induced EMF is given as;E = (ΔΦ) / t
Where,E = induced EMF in the loop
ΔΦ = change in magnetic flux through the loopt = time interval
So,ΔΦ = Φ₂ - Φ₁
Where,
Φ₂ = magnetic flux through the loop when the angle is 80°
Φ₁ = magnetic flux through the loop when the angle is 72°
Put the values in the above formula, we gget
ΔΦ = Φ₂ - Φ₁= (6.2 × 10⁻⁷) - (5.0 × 10⁻⁷) = 1.2 × 10⁻⁷ Wb (correct to two significant figure)
Now putting the values in the formula of induced EMF, we get;
E = (ΔΦ) / t= (1.2 × 10⁻⁷) / (0.5)= 2.4 × 10⁻⁷ V (correct to two significant figures)
Hence, the induced EMF in the loop is 2.4 × 10⁻⁷ V.
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12mm bar of hla meter gets stretched by 3 mm Under a Steady load of 8 KN what stress would be produced intbar by a wieght of soo N, which falls through 80 mm before commencing the stretching of the rod, which is tnitially unstressed E= LookN/mm²
The stress produced in the bar by the weight is 2,500 N/mm².
To calculate the stress produced in the bar, we need to consider the weight of 800 N that falls through a distance of 80 mm before the stretching of the rod begins. We can use Hooke's Law, which states that stress is directly proportional to strain, to find the stress.
The strain in the bar can be calculated using the formula:
Strain = Change in length / Original length
Given that the bar stretches by 3 mm and the original length is 12 mm, the strain can be calculated as:
Strain = 3 mm / 12 mm = 0.25
Now, we can use Hooke's Law to find the stress:
Stress = Young's modulus * Strain
Given that the Young's modulus (E) is 10,000 N/mm², we can calculate the stress:
Stress = 10,000 N/mm² * 0.25 = 2,500 N/mm²
Therefore, the stress produced in the bar by the weight is 2,500 N/mm².
It's worth noting that the steady load of 8 kN mentioned in the question does not affect the stress calculation since it acts after the stretching of the rod has already occurred. The weight falling through 80 mm is what causes the stretching and determines the stress in the bar.
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2.- A cylinder 8 cm in diameter and 60 cm long with surface temperature of 40 °C is placed in air at 15 °C flowing with velocity of 50 kw/h. What is the heat loss from the cylinder in W?
The heat-loss from the cylinder is 30.24 watts (W).
To calculate the heat loss from the cylinder, we can use the concept of convective heat transfer. The heat transfer rate can be determined using the following formula:
Q = h * A * ΔT
Where:
Q is the heat transfer rate (in watts, W)
h is the convective heat transfer coefficient (in W/m²·°C)
A is the surface area of the cylinder (in square meters, m²)
ΔT is the temperature difference between the surface of the cylinder and the surrounding air (in °C)
First, let's calculate the surface area of the cylinder. The surface area of the curved part (excluding the ends) can be calculated using the formula:
A_curved = π * D * L
Where:
D is the diameter of the cylinder (in meters, m)
L is the length of the cylinder (in meters, m)
Converting the given measurements to meters:
D = 8 cm = 0.08 m
L = 60 cm = 0.6 m
Calculating the surface area of the curved part:
A_curved = π * 0.08 m * 0.6 m
Next, we need to calculate the convective heat transfer coefficient, h.
The convective heat transfer coefficient depends on various factors such as the flow velocity, fluid properties, and geometry of the object. In this case, we are given the airflow velocity of 50 km/h.
To proceed further, we need to convert the airflow velocity to m/s:
Velocity = 50 km/h = (50 * 1000) m / (60 * 60) s
Next, we need to know the convective heat transfer coefficient associated with the given airflow velocity.
This coefficient depends on various factors and may require experimental or empirical data specific to the cylinder and airflow conditions.
In the absence of this information, let's assume a reasonable value for forced convection in air, such as h = 10 W/m²·°C.
With the obtained values, we can calculate the temperature difference (ΔT):
ΔT = 40 °C - 15 °C
Now, we can substitute the values into the formula to calculate the heat loss:
Q = h * A_curved * ΔT
Substituting the known values:
Q = 10 W/m²·°C * (π * 0.08 m * 0.6 m) * (40 °C - 15 °C)
Calculating the heat loss:
Q ≈ 10 W/m²·°C * (0.12096 m²) * 25 °C
Q ≈ 30.24 W
Therefore, the heat loss from the cylinder is approximately 30.24 watts (W).
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