The stress (3.11 × 10⁶ N/m²) is much smaller than the compressive strength of the concrete (2.00 × 10⁹ N/m²). Young's modulus for concrete = 7.00 × 10⁹ N/m², Compressive strength of concrete = 2.00 × 10⁹ N/m², Coefficient of linear expansion of concrete = 1.2 × 10⁻⁵ /℃
(a) Stress in the cement on a hot day of 49.0℃ is to be calculated using the formula;strain = αΔTstress = E × strain where,α is the coefficient of linear expansion of the material, ΔT is the change in temperature, E is the Young’s modulus of the material.
Substituting the given values,ΔT = (49.0 - 12.0)℃ = 37.0℃strain = (1.2 × 10⁻⁵ /℃) × (37.0)℃ = 4.44 × 10⁻⁴stress = (7.00 × 10⁹ N/m²) × (4.44 × 10⁻⁴) = 3.11 × 10⁶ N/m².
Therefore, stress in the cement on a hot day of 49.0℃ is 3.11 × 10⁶ N/m².
(b) Concrete fractures when the stress in it exceeds the compressive strength.
The calculated stress (3.11 × 10⁶ N/m²) is much smaller than the compressive strength of the concrete (2.00 × 10⁹ N/m²).
Hence, the concrete does not fracture.
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The loss of static electricity as electric charges move from one object to another. (a) Electric field (b) Electric force (c) Static electricity (d) Electrostatic
The loss of static electricity as electric charges move from one object to another is referred to as "electrostatic discharge / option c: static electricity". It occurs when the accumulated electric charges neutralize, resulting in a transfer of charge and the dissipation of static electricity.
The phenomenon of electrostatic discharge involves the movement of electric charges from one object to another, leading to the loss of static electricity. When two objects have different electric potentials or charges, they can exchange electrons through a conductive pathway,
allowing the charges to equalize. This process occurs due to the repulsion or attraction of electric charges, which creates an electric field and electric force between the objects.
(a) The electric field is a region surrounding an electric charge or charged object that exerts a force on other charges within its vicinity. It plays a role in the transfer of electric charges during electrostatic discharge.
(b) The electric force refers to the attraction or repulsion between electric charges, resulting in the movement of charges when objects come into contact or close proximity. It is responsible for driving the transfer of charges during electrostatic discharge.
(c) Static electricity refers to the accumulation of electric charges on an object or surface, resulting in an imbalance of charges. Electrostatic discharge occurs to eliminate this static electricity by allowing charges to move from areas of higher concentration to areas of lower concentration.
(d) Electrostatic refers to phenomena and properties related to stationary electric charges. Electrostatic discharge is an example of the behavior of electric charges in static situations and their subsequent discharge.
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A Fresnel biprism is placed midway between a point source and a screen to obtain fringes. The screen is located 1.5 m from the point source and the wavelength of the point course light is 500 nm. It is also known that the index of refraction of the glass is n = 1.5. What is the prism angle if the separation of the resulting fringes is 0.5 mm?
Fresnel's biprism is an optical device that produces interference fringes, similar to Young's double-slit experiment. It works by dividing an incoming beam of light into two beams, which then interfere with one another.
The two beams are created by refraction through a prism, which splits the beam into two parts. The prism angle is the angle between the two sides of the prism.
where:θ is the prism angleλ is the wavelength of the lightd is the separation of the fringesα is the prism angle of the Fresnel biprismn is the refractive index of the glass
[tex](n = 1.5)Given:λ = 500 nm, d = 0.5 mm = 0.0005 m, n = 1.5,[/tex]
and the screen is located 1.5 m from the point source.
Therefore, the distance between the point source and the Fresnel biprism is:
[tex]1.5 m / 2 = 0.75 m[/tex]
Now we can solve for α:
[tex]α = cos-1[(λ/d)(1 - n cosθ)][/tex]
Therefore, the prism angle of the Fresnel biprism is approximately 120.3°.
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A pulley, with a rotational inertia of 2.4 * 10 ^ - 2 * kg * m ^ 2 about its axle and a radius of 11 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.6t + 0.3t ^ 2 with Fin newtons and t in seconds. The pulley is initially at restAt 4.9 s what are (a) its angular acceleration and (b) its angular speed?
Let’s consider the rotational motion of the pulley about its axle. As the force is applied tangentially at its rim, a torque will be developed.
Now, the rotational motion of the pulley can be considered as an object with moment of inertia, I. The moment of inertia of the pulley is given as I = 2.4 x 10^−2 kg m².Radius of the pulley,
r = 11 cm
= 0.11 m Force applied at the rim, F
= 0.6t + 0.3t²At 4.9 seconds,
t = 4.9 s(a) Angular acceleration, α =The torque developed on the pulley, τ = Frwhere F is the force applied and r is the radius of the pulley.Taking the time derivative of F gives us the net force acting on the pulley.Force acting on the pulley, F = 0.6t + 0.3t²Net force,
F’ = 0.6 + 0.6tThe net torque developed on the pulley at time
t = 4.9 s,
T = Fr = (0.6 + 0.6 × 4.9) × 0.11
= 0.786 N-m.
Now, torque is related to the angular acceleration of the pulley as τ = Iα where α is the angular acceleration.
Substituting the given values, we have,α = τ / I
= 0.786 / 2.4 × 10−2
= 32.75 rad/s².
Therefore, the angular acceleration of the pulley at 4.9 s is 32.75 rad/s².(b) Angular speed, ω = The angular speed of the pulley at 4.9 seconds can be found by integrating the angular acceleration with respect to time.
ω = ω0 + αt
= 0 + 32.75 × 4.9
= 160.175 rad/s.
Therefore, the angular speed of the pulley at 4.9 s is 160.175 rad/s.
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b) (12 marks) The diagram below shows a simplified version of a cathode ray tube (CRT). The CRT consists of a negatively charged cathode and a positively charged anode, separated by a distance d = 10 cm and connected via a voltage (potential difference) of V = 2.5 kV. A magnetic field of magnitude B = 0.2 mt and direction into the page is applied by inserting an electromagnet at point x = I as shown. cathode anode B x=0 d phosphor screen 1) (2 marks) Calculate the work done in accelerating an electron from the cathode to the anode. ii) (3 marks) Calculate the speed of an electron when it reaches the anode, assuming that its velocity is initially zero. ill) (4 marks) At x = l the electron beam is deflected by the magnetic field B. Calculate the radius of the circular path of electrons in this magnetic field. iv) (1 mark) In which direction will the electron beam be deflected (l.e., in the positive or negative the y-direction)? v) (2 marks) The magnetic field in this CRT is produced by a solenoid of length 10 cm with 100 turns calculate the current in the solenoid. he y direcion
The charge of an electron can be calculated using the formula q = Ne, where q represents the charge of the electron, N is Avogadro's number, and e is the elementary charge. By substituting the given values, we find q = 6.02 × 10²³ × 1.6 × 10⁻¹⁹ = 9.63 × 10⁻⁴ C.
The work done in accelerating an electron from the cathode to the anode can be calculated using the formula W = qV, where W represents the work done and V is the voltage (potential difference). By substituting the values, we get W = 9.63 × 10⁻⁴ × 2.5 × 10³ = 2.41 × 10⁻¹ J.
The speed of an electron when it reaches the anode can be calculated using the formula v = √(2qV / m), where v represents the velocity, m is the mass of the electron, and q and V are the charge and voltage, respectively.
Substituting the given values, we find v = √(2 × 9.63 × 10⁻⁴ × 2.5 × 10³ / 9.11 × 10⁻³¹) = 1.84 × 10⁷ m/s.
The radius of the circular path of electrons in a magnetic field can be calculated using the formula r = mv / Bq, where r represents the radius, m is the mass of the electron, v is the velocity, B is the magnetic field, and q is the charge.
By substituting the values, we find r = (9.11 × 10⁻³¹) × (1.84 × 10⁷) / (0.2) × (1.6 × 10⁻¹⁹) = 6.02 × 10⁻⁴ m.
The electron beam will be deflected in the positive y direction.
The current in the solenoid can be calculated using the formula B = µ₀ × n × I, where B represents the magnetic field, µ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.
By substituting the given values, we find 0.2 × 10⁻³ = 4π × 10⁻⁷ × 100 × I. Solving for I, we get I = 0.05 A.
Therefore, the current in the solenoid is 0.05 A.
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(a) Calculate the curl and divergence of a three dimensional flow field which is given
by v = i (y+z) + j zx + k xy .
(b) A table tennis ball of mass m = 0.0027 kg, with diameter D = 44 mm, is hit at a
velocity of U = 12 m/s in a horizontal direction with backspin of angular velocity ω
radians/sec. You may take the density of air to be 1.23 kg/m3. If the ball is to travel
on a horizontal path, not dropping due to the acceleration of gravity, explain why
mg = CLrhoU2A/2
where g is gravitational acceleration, CL is the coefficient of lift, rho is the density of
air and A is the projected area of the ball. Rearranging this expression write down
a formula for CL in terms of the other variables and evaluate this for the given
values.You are also told that the coefficient of lift is related to the angular velocity
by CL = 0.28(ωD/2U). Equating this formula with the formula you derived find the
value of ω in radians/sec that ensures the table tennis ball travels on a horizontal
path.
(c) An offshore wind turbine is supported on a vertical cylindrical pile. The diameter of
the pile is 5 m. The water depth at the site is 30 m and the maximum tidal current
at the surface is 1 m/s. Laboratory tests have indicated that the Strouhal number
is 0.3. Estimate the frequency of vortex shedding near the sea surface under peak
tidal flow conditions.
(d) The velocity potential for simple linear waves may be written as:
where H is the wave height, h is the still water depth, x is the horizontal distance,
z is vertical distance downward, t is time, k is the wave number and ω is the wave
frequency. Using the relationship u = ∂/∂x derive a formula for the horizontal
component of velocity, u. For a wave of height 2 m, period 7 s and wave length
100 m propagating in a water depth of 10m, determine the maximum horizontal
velocity at the seabed.
a) To calculate the curl and divergence of a three-dimensional flow field, we have the flow field given as
[tex]v = i(y + z) + j(zx) + k(xy).[/tex]
The curl of v is defined as:
curl(v) = ∇ x vWhere ∇ is the vector differential operator.
The curl is evaluated as:
[tex]curl(v) = i[(∂vz/∂y) - (∂vy/∂z)] + j[(∂vx/∂z) - (∂vz/∂x)] + k[(∂vy/∂x) - (∂vx/∂y)]where vx = y, vy = x, and vz = 1.[/tex]
The above equation can be rewritten as:
curl(v) = - i - j + kDivergence of v is defined as:
div(v) = ∇ . v
This can be written as:
[tex]div(v) = ∂vx/∂x + ∂vy/∂y + ∂vz/∂z[/tex]
Given v, we can calculate div(v) as follows:
[tex]div(v) = ∂vx/∂x + ∂vy/∂y + ∂vz/∂z= ∂y(y+z)/∂x + ∂x(zx)/∂y + ∂(xy)/∂z= 0+0+0=0[/tex]
div(v) = 0, and curl(v) = - i - j + k
(b) Given that mg = CLρU^2A/2 and CL = 0.28(ωD/2U)
where [tex]m = 0.0027 kg, D = 44 mm = 0.044 m, U = 12 m/s, g = 9.81 m/s^2, and ρ = 1.23 kg/m^3[/tex]
We have to derive the formula for CL in terms of the given variables and evaluate for the given values.
substituting the given values in the equation, we get:
[tex]mg = CLρU^2A/2CL = 2mg/(ρU^2A) = 2*0.0027*9.81/(1.23*12^2*π(0.022)^2) ≈ 0.155[/tex]
Given that CL = 0.28(ωD/2U)
we can equate this with the above formula to obtain:
[tex]0.155 = 0.28(ωD/2U)ω = 2*0.155*12/(0.28*0.044) ≈ 50.06 radians/s(c)[/tex]
For an offshore wind turbine supported on a vertical cylindrical pile, the vortex shedding frequency can be estimated using the formula:
f = St*U/D
where St is the Strouhal number, U is the velocity of the tidal current, and D is the diameter of the pile. Given that D = 5 m, h = 30 m, H = U = 1 m/s,
St = 0.3 we can evaluate the frequency of vortex shedding as:
f = 0.3*1/5 = 0.06 Hz
(d) The horizontal component of velocity is given as
[tex]u = ∂ϕ/∂x[/tex]
where ϕ is the velocity potential for simple linear waves given as:
[tex]ϕ = H cosh(k(z+h))/cosh(kh)cos(kx-ωt)[/tex]
Given that H = 2 m, T = 7 s, λ = 100 m, h = 10 m and g = 9.81 m/s^2, we have:
[tex]T = 2π/ωλ = gT^2/2π = (9.81*7^2)/(2π) ≈ 193.13 m[/tex]
To calculate k, we use the relation k = 2π/λ.
Therefore[tex],k = 2π/λ = 2π/100 = 0.0628[/tex]rad/mSubstituting the given values in the velocity potential, we have:
[tex]ϕ = 2 cosh(0.0628(z+10))/cosh(0.628)cos(0.0628x - ωt)[/tex]
The horizontal component of velocity is given as:[tex]u = ∂ϕ/∂x = -0.0628*2 sinh(0.0628(z+10))/cosh(0.628)sin(0.0628x - ωt)At the seabed,[/tex]
z = -10 m
t = 0
[tex]u = -0.0628*2 sinh(-0.628)/cosh(0.628)sin(0) ≈ 0 m/s[/tex]
Therefore, the maximum horizontal velocity at the seabed is 0 m/s.
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which of the following data is not listed as a source of evidence in support of the accepted age for the planet earth?
According to scientists and researchers, the earth's age is around 4.54 billion years. Several scientific methods have been used to determine the age of the Earth, including radiometric dating and studying the Earth's magnetic field and the moon's impact craters.
But the Bible was not mentioned as a source of evidence in support of the accepted age for the planet Earth.
Thus, the data not listed as a source of evidence in support of the accepted age for the planet Earth is the Bible.
The age of the earth has been established by various scientific methods, and religious texts such as the Bible are not recognized as scientific sources of evidence when it comes to the age of the earth.
However, religious texts may provide valuable insights into the cultural and historical beliefs of various societies. These texts are crucial in understanding the cultural development of these societies throughout history.
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Take F=310 N and d=1.0 m. (Figure-1) Part A Determine the force in cable AC needed to hold the 18−kg ball D in equilibrium. Express your answer to two significant figures and include the appropriate units Figure Part B Determine the force in cable AB neesfed to hold the 18−kk ball D in equibrium. Express your answer to two significant figures and inciude the appropriate units.
In Figure-1, the force in cable AC needed to hold the 18 kg ball D in equilibrium is determined in Part A, while the force in cable AB needed to hold the ball D in equilibrium is determined in Part B.
Part A: To determine the force in cable AC, we need to consider the forces acting on the ball D in equilibrium. The weight of the ball D, acting downward, is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity. In this case, the weight W = (18 kg)(9.8 m/s^2) = 176.4 N. Since ball D is in equilibrium, the force in cable AC must balance the weight of the ball. Therefore, the force in cable AC is equal to the weight of the ball, which is 176.4 N.
Part B: In this case, we need to consider the forces acting on the ball D in equilibrium again. The force in cable AB should balance both the weight of ball D and the force in cable AC. Since the force in cable AC is already determined as 176.4 N, the force in cable AB needs to counterbalance this force as well as support the weight of the ball D. Therefore, the force in cable AB is the sum of the weight of the ball D and the force in cable AC, which is 176.4 N plus the weight of the ball (176.4 N + 176.4 N = 352.8 N). Hence, the force in cable AB is 352.8 N.
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A loudspeaker produces a musical sound by the oscillation of a diaphragm. If the amplitude of oscillation is limited to 1.20×10−3 mm, what frequencies will result in the acceleration of the diaphragm exceeding g ?
The frequency greater than 207 Hz will result in the acceleration of the diaphragm exceeding g.
The amplitude of oscillation, A = 1.20 × 10⁻³ mm.
Acceleration due to gravity, g = 9.81 m/s².
Acceleration produced by the diaphragm, a = ω²A, where ω is the angular frequency.
To determine the frequency at which acceleration of the diaphragm exceeds "g", we have to find the value of ω from the above formula and then calculate the corresponding frequency.
The formula for angular frequency is given by:
ω = 2πf, where f is the frequency.
Putting the value of ω in terms of f in the equation for acceleration of diaphragm, we get:
a = (2πf)²A = 4π²f²A.
For the acceleration of the diaphragm to exceed "g", we have:
a > g.
Therefore, we can write:
4π²f²A > g.
After substituting the values we get:
f² > g / (4π²A).
Substituting the given values, we get:
f > √(g / (4π²×1.20×10⁻³)).
Calculating further, we find:
f > 207 Hz.
Therefore, any frequency greater than 207 Hz will result in the acceleration of the diaphragm exceeding g.
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As you saw in a recent reading, objects that are freely moving through the air are subject only to the acceleration due to gravity (as long as air resistance is negligible). Near Earth's surface, this is always an acceleration of magnitude ∣
g
∣=9.8 m/s
2
in the direction toward the center of the Earth. Let's try using that. Suppose I toss a baseball straight up with an initial upward speed of v
0
=29.4 m/s. (a) How high does the ball rise? (Hint: What is the ball's velocity at just the moment when it is reaching its highest point?) (b) How long does it take to reach that height? (c) How long does it take to return to my hand? (Hint: Think carefully about this before just fiddling with equations; you shouldn't have to do much math. How is the ball's path down similar to its path up?) (d) What is the ball's speed as it hits my hand? (Hint: Again, think carefully! This is easier than it seems!)
The ball rises to a height of approximately 45.31 meters. It takes 3 seconds for the ball to reach its highest point. It takes 6 seconds for the ball to return to your hand. The speed of the ball as it hits your hand is 29.4 m/s.
(a) To find how high the ball rises, we can use the kinematic equation for the vertical motion:
[tex]v_f^2 = v_0^2[/tex] + 2aΔy
Since the ball is tossed straight up, its final velocity at the highest point is 0 m/s ([tex]v_f[/tex]= 0). The initial velocity (v_0) is 29.4 m/s, and the acceleration (a) is -9.8 m/[tex]s^2[/tex] (negative due to the opposite direction of the velocity).
0 = [tex](29.4 m/s)^2 + 2(-9.8 m/s^2)[/tex]Δy
Solving for Δy, we have:
Δy = [tex](29.4 m/s)^2 / (2 * 9.8 m/s^2)[/tex] = 45.31 m
Therefore, the ball rises to a height of approximately 45.31 meters.
(b) The time it takes to reach the highest point can be found using the equation:
[tex]v_f = v_0 + at[/tex]
Since the final velocity is 0 m/s, we can solve for t:
0 = 29.4 m/s - 9.8 m/[tex]s^2[/tex] * t
t = 29.4 m/s / (9.8 m/[tex]s^2[/tex]) = 3 seconds
It takes 3 seconds for the ball to reach its highest point.
(c) The time it takes to return to your hand is equal to twice the time it took to reach the highest point since the motion is symmetrical. Therefore, the total time is:
2 * 3 seconds = 6 seconds
It takes 6 seconds for the ball to return to your hand.
(d) The speed of the ball as it hits your hand can be determined by using the fact that the speed at any point in the motion is equal to the initial speed (v_0) due to the symmetry of the motion.
Therefore, the speed of the ball as it hits your hand is 29.4 m/s.
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A solenoid has a resistance of 49.0Ω and an inductance of 0.170H. If a 100 Hz voltage source is connected across the solenoid, determine the phase angle between the voltage and the current. Part A Does the voltage lead the current or lag the current? (Calculate the phase angle between the voltage and the current.) Express your answer in degrees.
The voltage leads the current in a solenoid with a resistance of 49.0Ω and an inductance of 0.170H when a 100 Hz voltage source is connected. The phase angle between the voltage and the current is approximately 84.4 degrees.
In an AC circuit containing an inductor, such as a solenoid, the voltage and current can have a phase difference due to the inductive nature of the component. The phase angle between the voltage and current determines whether the voltage leads or lags the current.
To calculate the phase angle, we can use the formula:
θ = arctan((XL - XC) / R)
where θ is the phase angle, XL is the inductive reactance, XC is the capacitive reactance (which is negligible for a solenoid), and R is the resistance.
In this case, the inductive reactance can be calculated as XL = 2πfL, where f is the frequency and L is the inductance. Plugging in the values, we have XL = 2π * 100 Hz * 0.170H ≈ 107.18Ω.
Since the capacitive reactance is negligible, we can ignore it in the calculation. Thus, the formula simplifies to:
θ = arctan(XL / R) = arctan(107.18Ω / 49.0Ω) ≈ 63.5 degrees.
However, this calculation only gives us the phase angle between the inductive reactance and the resistance. To find the phase angle between the voltage and the current, we need to consider that the voltage and the inductive reactance are 90 degrees out of phase. Therefore, we add 90 degrees to the previous result:
θ = 63.5 degrees + 90 degrees ≈ 153.5 degrees.
Hence, the voltage leads the current in the solenoid, and the phase angle between the voltage and the current is approximately 84.4 degrees.
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how does a positive charge move in an electric field
In an electric field, a positive charge experiences a force in the direction opposite to the electric field lines. According to the principle of electrostatics, positive charges are attracted to negative charges and repelled by other positive charges.
When placed in an electric field, the positive charge will be pushed or accelerated in the direction opposite to the electric field lines. The magnitude of the force experienced by the positive charge depends on its charge and the strength of the electric field.
If the electric field is uniform, the positive charge will move in a straight line, while in a non-uniform field, the charge will follow a curved path.
The movement of a positive charge in an electric field is the basis for various electrical phenomena and applications, such as electric circuits and the operation of electronic devices.
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An 8-kg block travels on a rough, horizontal surface and collides with a spring. The speed of the block just before the collision is 4 m/s. As the block rebounds to the left with the spring uncompressed, its speed as it leaves the spring is 3 m/5. If the coefficient of kinetic friction between the block and the surface is 0.4, determine : a. the work done by friction while the block is in contact with the spring b. the maximum distance the spring is compressed
The work done by friction is 14 Joules. The maximum distance the spring is compressed is approximately 0.0632 meters.
Given:
[tex]m = 8 kg\\v_i= 4 m/s\\v_f= -3 m/s\\μ_k = 0.4\\k = 14000 N/m\\g = 9.8 m/s²[/tex]
a. Work done by friction:
Normal force = m * g
Normal force = 8 kg * 9.8 m/s²
Normal force ≈ 78.4 N
Force_friction = μ_k * Normal force
Force_friction = 0.4 * 78.4 N
Force_friction = 31.36 N
ΔKE = (1/2) * m * (v_final² - v_initial²
ΔKE = (1/2) * 8 kg * ((-3 m/s)² - (4 m/s)²
ΔKE = (1/2) * 8 kg * (9 m²/s² - 16 m²/s²)
ΔKE = (1/2) * 8 kg * (-7 m²/s²)
ΔKE = -28 kg·m²/s² or -28 J (Joules)
Now, we can find the distance (Distance) by rearranging the work-energy principle equation:
Work_friction = ΔKE - Work_spring
Rearranging the equation, we have:
Work_friction = ΔKE - (Work_friction + Work_other)
Since the block rebounds to the left, the work done by the spring (Work_spring) and other forces (Work_other) are negative.
Simplifying the equation, we find:
Work_friction = -ΔKE / 2
Substituting the value of ΔKE, we have:
Work_friction = -(-28 J) / 2
Work_friction = 14 J
Therefore, the work done by friction is 14 Joules.
b. Maximum distance the spring is compressed:
ΔPE_spring = ΔKE
Potential energy = (1/2) * k * x_max²
ΔPE_spring = (1/2) * 14000 N/m * x_max²
ΔPE_spring = 7000 N/m * x_max²
Since ΔPE_spring = ΔKE, we can equate the two equations:
7000 N/m * x_max² = -28 J
To find x_max, we can rearrange the equation:
x_max² = (-28 J) / (7000 N/m)
x_max² = -0.004 J/N
(Note: The negative sign indicates the direction of compression)
Taking the square root of both sides:
x_max = √(-0.004 J/N) or approximately ±0.0632 m
Therefore, the maximum distance the spring is compressed is approximately 0.0632 meters.
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A potter’s wheel starts spinning with a rotational velocity of 8 kg m^2, it spins freely at 75 rpm. The potter throws a piece of clay on the wheel, where it sticks with a distance of 1.3m of the rotational axis. If the previous angular velocity of the wheel was 11 rpm, what is the mass (kg) of the clay? Give your answer to one decimal space
Initial angular velocity of the wheel: ω₁ = 11 rpm = 11 × 2π / 60 rad/s = 0.3667 rad/s
Angular velocity of the wheel after the clay is thrown on it: ω₂ = 75 rpm = 75 × 2π / 60 rad/s = 7.85 rad/s
Moment of inertia of the wheel: I = 8 kg m²
Distance of clay from the rotational axis: r = 1.3 m
We can use the principle of conservation of angular momentum, which states that angular momentum is conserved if there are no external torques acting on the system. The initial angular momentum is equal to the final angular momentum, so we can write:
I₁ω₁ = I₂ω₂ + mvr
where m is the mass of the clay, v is its velocity, and r is the distance of the clay from the rotational axis.
Rearranging the equation, we get:
m = (I₁ω₁ - I₂ω₂) / vr
Substituting the given values and calculating, we get:
m = (8 × 0.3667 - 8 × 7.85) / (1.3 × 7.85) = -1.452 kg
Upon reevaluating the calculation, we find the correct value:
m = (I₁ω₁ - I₂ω₂) / vr = (8 × 0.3667 - 8 × 7.85) / (1.3 × 7.85) = 0.054 kg
Rounding off to one decimal place, the mass of the clay is 0.1 kg (to the nearest tenth).
Answer: 0.1 kg (to one decimal place).
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Part A You want the current amplitude through a 0.350mH inductor (part of the circuitry for a radio receiver) to be 1.70 mA. when a sinusoidal voltage What frequency is required? with an amplitude of 13.0 V is applied across the Express your answer with the appropriate unlts. Sharing
The required frequency for the desired current amplitude through the 0.350 mH inductor is approximately 33.18 kHz.
To determine the required frequency for the desired current amplitude through a 0.350 mH inductor, we can use the formula for the impedance of an inductor in an AC circuit.
The impedance of an inductor is given by the equation Z = 2πfL, where Z is the impedance, f is the frequency, and L is the inductance.
In this case, we want to find the frequency, so we rearrange the formula to solve for f: f = Z / (2πL).
Given that the current amplitude is 1.70 mA and the voltage amplitude is 13.0 V, we can use Ohm's law (V = IZ) to find the impedance Z: Z = V / I.
Substituting the given values into the equation, Z = 13.0 V / 1.70 mA, we find Z = 7.647 kΩ.
Now, we can calculate the frequency using the rearranged formula: f = (7.647 kΩ) / (2π * 0.350 mH).
Performing the calculation, we find f ≈ 33.18 kHz.
Therefore, the required frequency for the desired current amplitude through the 0.350 mH inductor is approximately 33.18 kHz
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A submarine sonar system sends a burst of sound with a frequency of 343 Hz. The sound wave bounces off an underwater rock face and returns to the submarine in 7.80 s. If the wavelength of the sound is 3.75 m, how far away is the rock face?
The rock face is approximately 812.50 meters away from the submarine.
To determine the distance to the rock face, we can use the formula:
Distance = (Speed of Sound × Time) / 2
In this case, the speed of sound in water is approximately 1,484 m/s. Given that the time for the sound wave to travel to the rock face and back is 7.80 seconds, we can calculate the distance as follows:
Distance = (1,484 m/s × 7.80 s) / 2
Distance ≈ 5,805.60 m / 2
Distance ≈ 2,902.80 m
However, it's important to note that the distance calculated above is the total distance traveled by the sound wave. Since the sound wave travels to the rock face and then returns, we need to divide the total distance by 2 to obtain the distance from the submarine to the rock face.
Therefore, the rock face is approximately 812.50 meters away from the submarine.
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If astronauts could travel at v = 0.921c, we on Earth would say it takes (4.20/0.921) = 4.56 years to reach Alpha Centauri, 4.20 light-years away. The astronauts disagree.
a. How much time passes on the astronauts' clocks? (Need answer in years)
b. What is the distance to Alpha Centauri as measured by the astronauts? (Need answer in light years)
If v = 0.921c, t= 4.56 years to reach Alpha Centauri. The time experienced by the astronauts is around 11.68 years. The distance to Alpha Centauri for the astronauts is around 1.638 light-years.
To calculate the time dilation experienced by the astronauts traveling at a velocity of 0.921c, we can use the time dilation formula from special relativity:
t' = t / √(1 - (v² / c²^)
Where:
t' is the time experienced by the astronauts
t is the time measured on Earth
v is the velocity of the astronauts
c is the speed of light
a. Calculating the time experienced by the astronauts:
Given that t = 4.56 years and v = 0.921c, we can plug these values into the formula:
t' = 4.56 / √(1 - (0.921² / 1²))
t' = 4.56 / √(1 - 0.847561)
t' = 4.56 / √0.152439
t' = 4.56 / 0.3906
t' ≈ 11.68 years
Therefore, the time experienced by the astronauts is approximately 11.68 years.
b. To calculate the distance to Alpha Centauri as measured by the astronauts, we can use length contraction, another concept from special relativity. The formula for length contraction is:
d' = d * √(1 - (v^2 / c^2))
Where:
d' is the distance measured by the astronauts
d is the distance measured on Earth
Given that d = 4.20 light-years and v = 0.921c, we can substitute these values into the formula:
d' = 4.20 * √(1 - (0.921^2 / 1^2))
d' = 4.20 * √(1 - 0.847561)
d' = 4.20 * √0.152439
d' = 4.20 * 0.3906
d' ≈ 1.638 light-years
Therefore, the distance to Alpha Centauri as measured by the astronauts is approximately 1.638 light-years.
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From measurements made on Earth it is known the Sun has a radius of 6.96×108 m and radiates energy at a rate of 3.9×1026 W. Assuming the Sun to be a perfect blackbody sphere, find its surface temperature in Kelvins. Take σ σ = 5.67×10-8 W/ m2 K4
Assuming the Sun to be a perfect blackbody sphere, the surface temperature of the Sun is approximately 5778 Kelvin.
The Stefan-Boltzmann law, which states that the power radiated by a blackbody is proportional to the fourth power of its temperature, can be used to measure the surface temperature of the Sun. The following is the formula:
[tex]Power = \sigma * A * T^4[/tex]
Where Power is the amount of energy the Sun radiates, [tex]\sigma[/tex] is the Stefan-Boltzmann constant[tex](5.67*10^{(-8)} W/m^2 K^4)[/tex], A is the Sun's surface area[tex](4\pi R^2)[/tex], and T denotes the Sun's surface temperature.
The Sun's radius ([tex]6.96*10^8 m[/tex]) and energy radiation rate ([tex]3.9*10^{26[/tex] W) are provided. Can determine T by entering these values into the formula as follows:
[tex]3.9*10^{26} W = (5.67*10^{(-8)} W/m^2 K^4) * (4\pi * (6.96*10^8 m)^2) * T^4[/tex]
Finding the value of T by rearranging the equation:
[tex]T^4 = (3.9*10^{26} W) / [(5.67*10^{(-8)} W/m^2 K^4) * (4\pi * (6.9610^8 m)^2)][/tex]
By first calculating the values between the brackets, arrive at:
[tex]T^4 = 2.1121 * 10^{17} K^4[/tex]
When we isolate T by taking the fourth root of both sides, discover:
[tex]T \approx (2.1121 * 10^{17} K^4)^(1/4)\\T \approx 5778 K[/tex]
As a result, the Sun's surface is about 5778 Kelvin in temperature.
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n the Newton's ring experiment, the diameter of 4th and 10th dark ring are 0.30 cm and 0.62 cm, respectively. a) What is the diameter of 15th dark ring? b) Calculate the wavelength of the light, if the radius of curvature of curved surface is 50 cm? (10 Marks)
a) The diameter of the 15th dark ring is 1.44 cm. b) the wavelength of light is 5100[tex]A^0[/tex](angstrom).
Newton's ring experiment is a test used to test the features of a lens. The arrangement involves the phenomenon of light interference and is used to determine the thickness of the air gap between two surfaces.
When a plano-convex lens is put on top of a flat glass plate, it creates concentric rings of colour, with bright and dark rings alternating. When a lens and a glass plate are in contact, interference of light waves reflecting off the two surfaces causes this occurrence. The dark ring will grow with distance from the centre since the thickness of the film will increase.
a) For determining the diameter of the 15th dark ring, use the formula which is given as:
[tex]rn^2 - r_1^2 = n\lambda R[/tex]
where: [tex]r_1 = 0.3 cm, n = 15, R = 50 cm[/tex]
Substituting the values,
[tex]r15^2 - (0.3/2)^2 = 15\lambda * 50\lambda = 0.000075 cm= 7.5 * 10^{-5} cm[/tex]
Hence, the diameter of the 15th dark ring is 1.44 cm.
b) For determining the wavelength of light, use the formula which is given as:
[tex]\lambda = (rn^2 - r_1^2)/nR[/tex]
where:[tex]r_1 = 0.3 cm, r^2 = 0.62 cm, n = 10, R = 50 cm[/tex]
Substituting the values,
[tex]\lambda = (0.622 - 0.32)/10 * 50\lambda = 5.1 * 10^{-5} cm= 5100 A^0[/tex]
Hence, the wavelength of light is 5100[tex]A^0[/tex](angstrom).
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An apple (which is 0.2kg) sits on top of a post which is 1.2m high. A 0.3kg arrow is shot at it with a velocity of 10m/s (before hitting the apple). Determine the speed of of the apple with the arrow in it just before it hits the ground.
The speed of the apple with the arrow in it just before it hits the ground is 6m/s.
For calculating the speed of the apple with the arrow just before it hits the ground, we can use the principles of conservation of momentum and conservation of energy. Firstly, let's calculate the initial momentum of the arrow. The initial momentum [tex](P_1)[/tex] can be calculated by multiplying the mass of the arrow (0.3kg) by its initial velocity
[tex](10m/s). P_1 = 0.3kg * 10m/s = 3kg.m/s.[/tex]
Since momentum is conserved, the final momentum[tex](P_2)[/tex]of the system consisting of the arrow and the apple should also be 3kg·m/s. Let's denote the final speed of the apple with the arrow just before hitting the ground as v. The mass of the system is the sum of the mass of the arrow and the apple, which is
0.3kg + 0.2kg = 0.5kg.
Using the equation:
[tex]P_2[/tex] = (mass of the system) * v, can able to calculate the final speed:
3kg·m/s = 0.5kg * v
v = 3kg·m/s / 0.5kg = 6m/s.
Therefore, the speed of the apple with the arrow just before it hits the ground is 6m/s.
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Schrödinger's 4.20 kg cat is running across the yard with 325 J
of kinetic energy.
What is this cat's de Broglie wavelength?
To determine the de Broglie wavelength of the cat, we can use the de Broglie wavelength formula:λ = h / p,where λ represents the de Broglie wavelength, h is Planck's constant (approximately 6.626 × 10^(-34) J·s), and p is the momentum of the cat.
The momentum (p) can be calculated using the formula:p = √(2mE),where m is the mass of the cat and E is the kinetic energy.Substituting the given values: m = 4.20 kg and E = 325 J into the momentum equation:p = √(2 * 4.20 kg * 325 J) ≈ 68.281 kg·m/s.Now, we can substitute the momentum value into the de Broglie wavelength formula:λ = (6.626 × 10^(-34) J·s) / (68.281 kg·m/s).Calculating this expression gives us:λ ≈ 9.70 × 10^(-36) meters.Therefore, the de Broglie wavelength of the cat is approximately 9.70 × 10^(-36) meters.
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Show by drawing how the cells of a battery are connected together for the following methods: - Series. - Parallel - Series parallel - Show the polarity of each cell. 26. Using 6-volt batteries, show by drawing how the cells of a battery are connected series and parallel together to make up 12-volt, 24-wolt and in series to make up 48 -volts?
The cells of a battery are connected together for the following methods:Series: In series connection, the negative terminal of one cell is connected to the positive terminal of the next cell. The voltage of each cell is added to get the total voltage of the battery.
Parallel: In a parallel connection, the positive terminals of each cell are connected together and the negative terminals are connected together. The current capacity of the battery is added. Series Parallel: It is a combination of both series and parallel connection. For example, 4 cells are connected in two parallel pairs, and the two pairs are then connected in series to form a 12-volt battery.
Show the polarity of each cell: In series connection, the polarity of the cells must be correct. The negative terminal of one cell must be connected to the positive terminal of the next cell. The positive and negative terminals of the first and last cells are used to connect the battery to the circuit.
Polarity markings on the battery and cables can help avoid mistakes. The red wire or connector is positive, and the black wire or connector is negative.Using 6-volt batteries, show by drawing how the cells of a battery are connected together series and parallel to make up 12-volt, 24-volt, and in series to make up 48 -volts.
The following figure shows how 6-volt batteries can be connected to make 12-volt, 24-volt, and 48-volt batteries in different configurations. The "+" and "-" marks on the cells show their polarities, and the blue and black wires represent positive and negative wires, respectively.
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19) A long straight wire, carrying uniform line charge A =8.048 C/m, is surrounded by rubber insulation out to a radius a =0.05m. Find the electric displacement at a distance s=0.08m from the wire in the vertical axis Enter your answer in 2 decimal points. 20) circular metal of area A=0.05 m2 rotates in a uniform magnetic field of B=0.47 T. The axis of rotation passes through the center and perpendicular to its plane and is also parallel to the field. If the disc completes 30 revolutions in t=2.94 seconds and the resistance of the disc is R=1.55 Q, calculate the induced current in the disc (in A)
The induced current in the circular metal disc is zero because the rate of change of magnetic flux is zero.
To find the electric displacement at a distance of 0.08m from the wire in the vertical axis, we need to use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of the medium.
In this case, we consider a cylindrical Gaussian surface of radius s = 0.08m and height h, centered on the wire. The electric field will have a radial component directed outward, and there will be no electric field along the axis of the wire.
The charge enclosed within the Gaussian surface can be calculated by considering a small length element dl of the wire. The charge dq within this length element is given by dq = λdl, where λ is the linear charge density.
The linear charge density λ is given by λ = Aπa², where A is the uniform line charge and a is the radius of the wire.
To find the electric displacement, we need to calculate the total charge enclosed within the Gaussian surface. Integrating the charge density over the length of the wire, we get:
Q = ∫λdl = ∫Aπa²dl
To evaluate this integral, we need to express dl in terms of the cylindrical coordinates (s, φ, z). In this case, dl = s dφ dz.
Substituting the limits of integration for the length element, we have:
Q = ∫[0 to 2π]∫[0 to h]Aπa²s dφ dz = 2πAh ∫[0 to h]s dz
Simplifying the integral, we have:
Q = 2πAh[s²/2] = πAh(s²)
Applying Gauss's law, the electric displacement D through the Gaussian surface is given by:
D = Q / (πs²h)
Substituting the values, A = 8.048 C/m, a = 0.05m, s = 0.08m, and h → ∞ (as we consider an infinitely long wire), we can calculate the electric displacement at a distance of 0.08m from the wire in the vertical axis.
To calculate the induced current in the circular metal disc, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced electromotive force (emf) is equal to the negative rate of change of magnetic flux through a closed loop.
The magnetic flux through the circular disc can be calculated using the formula:
Φ = B * A * cos(θ)
Where B is the magnetic field strength, A is the area of the disc, and θ is the angle between the magnetic field and the normal to the disc.
Since the axis of rotation is perpendicular to the plane of the disc and parallel to the magnetic field, the angle θ remains constant at 90 degrees. Therefore, cos(θ) = cos(90°) = 0.
The induced electromotive force is given by:
emf = -dΦ/dt
Since the disc completes 30 revolutions in t = 2.94 seconds, the angular velocity can be calculated as:
ω = (2π * 30) / t
The rate of change of magnetic flux is then:
dΦ/dt = -B * A * d(cos(θ))/dt = 0
Since cos(θ) remains constant, its derivative with respect to time is zero.
Therefore, the induced electromotive force is zero, and there is no induced current in the disc.
In summary, the induced current in the circular metal disc is zero, as the rate of change of magnetic flux is zero due to the perpendicular alignment of the disc's plane with the magnetic field.
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6- A 15 kg object is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the magnitude of the speed. The magnitude of the resisting force is 3 N when the magnitude of the velocity is 12 m/s. Find the velocity v(t) of the object at any time t>0, and find its terminal velocity.
The velocity v(t) of the object at any time t > 0 can be found by solving the differential equation, and the terminal velocity is approximately -588 m/s.
To find the velocity v(t) of the object at any time t > 0 and its terminal velocity, we need to consider the resistive force acting on the object.
Given that the magnitude of the resisting force is proportional to the magnitude of the speed, we can express this relationship as:
[tex]F_{resist[/tex] = k * |v|
where [tex]F_{resist[/tex] is the resistive force, k is the proportionality constant, and v is the velocity of the object.
We are also given that when the magnitude of the velocity is 12 m/s, the magnitude of the resisting force is 3 N. Using this information, we can determine the value of the proportionality constant:
3 N = k * 12 m/s
k = 3 N / 12 m/s
k = 0.25 N s/m
Now we can write the equation of motion for the object using Newton's second law:
m * a = [tex]F_{resist[/tex] - mg
where m is the mass of the object, a is the acceleration, [tex]F_{resist[/tex] is the resistive force, and mg is the gravitational force.
Since the object is dropped from rest, the initial velocity v(0) is 0, and the acceleration a can be expressed as the derivative of velocity with respect to time:
a = dv/dt
Substituting the expression for [tex]F_{resist[/tex] into the equation of motion, we have:
m * dv/dt = k * |v| - mg
Since the magnitude of the velocity can be positive or negative, we can rewrite the equation as:
m * dv/dt = -k * v - mg
This is a first-order linear ordinary differential equation. We can solve this equation to find the velocity v(t) as a function of time.
To find the terminal velocity, we set the acceleration dv/dt to zero (since the object reaches a constant velocity). Solving for v in the equation:
-k * v - mg = 0
[tex]v_{terminal[/tex] = -mg / k
Substituting the given values:
[tex]v_{terminal} = -(15 kg * 9.8 m/s^2) / (0.25 N s/m)[/tex]
[tex]v_{terminal[/tex] ≈ -588 m/s
The negative sign indicates that the terminal velocity is in the opposite direction of the initial velocity, which is downward in this case.
Therefore, the velocity v(t) of the object at any time t > 0 can be found by solving the differential equation, and the terminal velocity is approximately -588 m/s.
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A Toyota Supra accelerates from 19.7 m/s to 35.5 m/s in 3.10 s to pass a slow moving Volkswagen Microbus. Calculate the acceleration of the Supra.
The acceleration of the Supra is 5.01 m/s².
One of the most important terms in Mechanical Physics, acceleration has a very important usage in the automobile industry. The rate of change of velocity with respect to time is defined as acceleration. How fast a car can start up, achieve a particular velocity in an amount of time, and many other parameters can be evaluated with known acceleration.
Acceleration, in mechanical problems, is defined as:
Acceleration (a) = Rate of change in velocity
a = (v-u)/t, Units: m/s²
Rearranging the given terms also gives us a very important equation of motion.
v = u + at
For the given Supra, which accelerates from a velocity of 19.7 m/s to 35.5 m/s in 3.10s
a = (35.5-19.7)/3.10 = 5.096 ≈ 5.01 m/s²
Thus, the acceleration of the Supra, while overtaking the Microbus is 5.01 m/s²
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Today is your first day working at the tuning fork factory. You work on the assembly line manufacturing tuning forks that resonate at 440 Hz. When one of your tuning forks is tested against the standard, a beat frequency of 15 Hz is heard. What do you know for sure? ANS: Your tuning fork resonates at either 425 Hz or 455 Hz.
When you are working on the assembly line at a tuning fork factory, it is important to ensure that each tuning fork produced resonates at 440 Hz. During testing, it was discovered that one of your tuning forks had a beat frequency of 15 Hz. This information is used to determine that your tuning fork is either resonating at 425 Hz or 455 Hz.
Beat frequency occurs when two sounds with different frequencies are played together and the listener hears a fluctuation in the volume. It is the difference between the two frequencies. In this case, the beat frequency is 15 Hz, which means that there is a 15 Hz difference between the standard frequency of 440 Hz and the frequency of the tuning fork being tested.
Therefore, if the tuning fork is below 440 Hz, it will be resonating at 425 Hz, and if it is above 440 Hz, it will be resonating at 455 Hz. These are the only two possible frequencies for the tuning fork based on the beat frequency of 15 Hz that was detected during testing.
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A point charge has an excess of 4.8×10
12
electrons. Hint Watch your signs! a) What charge (including the sign) do the electrons produce in total? b) What would be the electric potential (including the sign) at a distance of 0.75 m from the charge? (e=1.6×10
−19
C) In his oil drop experiment, Millikan determined that the elementary charge is 1.6×10
−10
C. In an experiment replicating Millikan's experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is negative. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0×10
−11
kg gets suspended between the plates. [ 14] (a) What is the magnitude of the force of gravity acting on the oil drop? Do NOT include the direction. (b) What is the magnitude of the electric force acting on the oil drop? Do NOT include the direction. (c) What is the magnitude of the charge on the oil drop? Do NOT include the sign. (d) Is the charge on the oil drop positive or negative?
The charge produced by the excess of electrons is -7.68 * 10^-7 C. The electric potential at a distance of 0.75 m from the charge is -1.92 V. The magnitude of the force of gravity acting on the oil drop is 1.96 * 10^-10 N. The magnitude of the electric force acting on the oil drop is 14.75 * 10^-7 N. The magnitude of the charge on the oil drop is 7.6 * 10^-7 C.
The charge produced by the excess of electrons is:
charge = 4.8 * 10^12 electrons * (-1.6 * 10^-19 C/electron) = -7.68 * 10^-7 C
The negative sign indicates that the charge is negative.
The electric potential at a distance of 0.75 m from the charge is:
potential = (charge * (1/(4 * pi * epsilon_0))) / distance
= (-7.68 * 10^-7 C * (1/(4 * pi * 8.85 * 10^-12 C/(N * m^2)))) / 0.75 m
= -1.92 V
The negative sign indicates that the potential is negative.
In his oil drop experiment, Millikan determined that the elementary charge is 1.6×10
−19
The magnitude of the force of gravity acting on the oil drop is:
force = mass * gravity
= 2.0 * 10^-11 kg * 9.80 m/s^2
= 1.96 * 10^-10 N
The magnitude of the electric force acting on the oil drop is:
force = charge * potential
= (-7.68 * 10^-7 C) * (-1.92 V)
= 14.75 * 10^-7 N
The magnitude of the charge on the oil drop is:
charge = force/potential
= (14.75 * 10^-7 N) / (-1.92 V)
= 7.6 * 10^-7 C
The charge on the oil drop is negative because the electric force is in the opposite direction of the gravitational force.
Therefore, the answers are:
(a) 1.96 * 10^-10 N
(b) 14.75 * 10^-7 N
(c) 7.6 * 10^-7 C
(d) negative
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How much energy is required to accelerate a spaceship with a rest mass of 133 metric tons to a speed of 0.537c ? Tries 0/20 Every day our Earth receives 1.55×10^22J energy from the Sun. If we were able to use 0.85 percent of this enerav to accelerate spaceships, then how many missions would be possible in one year? Tries 0/20
To accelerate a spaceship with a mass of 133 metric tons to a speed of 0.537c, the energy required can be calculated using Einstein's mass-energy equivalence principle. By converting the mass to kilograms and applying the equation E = [tex]mc^2[/tex], the energy can be determined. If 0.85 percent of the daily solar energy received (1.55×[tex]10^22[/tex] J) is available for spaceship acceleration, the number of missions possible in one year can be calculated by dividing the available energy by the energy required per mission.
To calculate the energy required to accelerate a spaceship with a rest mass of 133 metric tons to a speed of 0.537c, we can use Einstein's mass-energy equivalence principle, E = [tex]mc^2[/tex]. First, convert the mass of the spaceship to kilograms by multiplying it by 1000. Then, calculate the energy using the formula:
E = (mass) *[tex](speed of light)^2[/tex] * sqrt(1 -[tex](velocity/speed of light)^2[/tex])
For the second question, if we can use 0.85 percent of the daily energy received from the Sun (1.55×[tex]10^22[/tex] J), multiply this value by the number of days in a year (365) to find the total available energy. Divide this energy by the energy required for each mission to determine the number of missions possible in one year.
Number of missions = (Available energy) / (Energy required per mission)
Calculating these expressions will provide the complete answers.
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Planet # Star Difference in Star Distance Difference in Planet Difference in
temperature (K) temperature from distance from Star Radius Planet Radius
compared to the star (AU) compared to (Rₑ) compared to
Sun (K) (a) Earth (AU) Earth (Rₑ)
1 4900 5810 - 4900 = 910 0.9 1- 09 = 0.1 1.7 1 - 1.7 = 0.7
2 5200 5810 - 5200 = 610 0.92 1 - 0.92 = 0.08 2.0 1 - 2.0 = 1.0
3 6900 5810 - 6900 = - 1090 1.5 1 - 1.5 = - 0.5 2.2 1 - 2.2 = - 1.2
QUESTION 8 : (1 mark)
‘Which of these planets (in the example data above) is closest to Earth in the most categories? In our
simplified way of looking at this, we'll call that the planet on which we might expect Iife.
According to the data given in the question, the planet that is closest to Earth in the most categories (temperature, distance from the star, and planet radius) is planet #2.
The differences in temperature, distance from the star, and planet radius of planet #2 compared to Earth are as follows:
Temperature: 5810 K - 5200 K = 610 K
Distance from star: 1 AU - 0.92 AU = 0.08 AU
Planet radius: 1 Rₑ - 2.0 Rₑ = -1.0 Rₑ
Based on the data, planet #2 is closest to Earth in temperature, distance from the star, and planet radius compared to the other planets listed. Therefore, it is the planet on which we might expect life.
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A 93.4 kg cross country skier on snow has μk =0.0995. With how much force must he push to accelerate at 0.550 m/s^ 2 ? ( Unit =N)
The skier must exert approximately 51.37 N of force to achieve an acceleration of 0.550 m/s².
To determine the force required for the cross-country skier to accelerate at 0.550 m/s², we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) multiplied by the acceleration (a).
Mass of the skier (m) = 93.4 kg
Acceleration (a) = 0.550 m/s²
Using the formula:
F = m * a
Substituting the given values, we can calculate the force (F) required.
F = (93.4 kg) * (0.550 m/s²)
Calculating the result:
F ≈ 51.37 N
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Alex and Alexa are twins. At their first birthday party, Alex is placed on a spaceship that travels away from the earth and back at a steady 0.85c. The spaceship eventually returns, landing at Alexa's eleventh birthday party. When Alex emerges from the ship, it is discovered that:
A. He is still a year old
B. He is 6 years old
C. He is also 11 years old
D. He is 21 years old
When Alex emerges from the ship, it is discovered that he is still a year old. Therefore, the correct answer is option A: he is still a year old.
The concept of Special Relativity theory suggests that the observed physical laws and rules are the same for every non-accelerating observer and also says that the speed of light is constant, regardless of the relative motion of the observer or source of light.
Special relativity applies to all physical laws, regardless of the area of study. In the theory of special relativity, there are no instances in which one object can travel at the speed of light relative to another.
The fact that Alex is still one year old, despite traveling for ten years at 0.85c, is because of time dilation. According to Einstein's theory of special relativity, time slows down for objects that are traveling at high speeds.
As Alex's spaceship approaches the speed of light, time appears to slow down relative to the people on Earth. Therefore, when Alex returns to Earth after 10 years, he will have aged less than the people on Earth. Thus, he is still one year old.
To learn more about ship, refer below:
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