Answer:
[tex]V = 900\; {\rm V}[/tex] for a resistor where [tex]I = 30\; {\rm A}[/tex] and [tex]R = 30\; {\rm \Omega}[/tex].
Explanation:
By Ohm's Law, if the electric resistance of a resistor is [tex]R[/tex] and the current going through that resistor is [tex]I[/tex], the voltage across that resistor would be [tex]V = I\, R[/tex].
Note that electric resistance is typically measured in Ohms, where [tex]1\; {\rm \Omega} = 1\; {\rm V \cdot A^{-1}}[/tex].
Given that [tex]I = 30\; {\rm A}[/tex] and [tex]R = 30\; {\rm \Omega}[/tex] for the resistor in this question, apply the formula [tex]V = I\, R[/tex] to find the voltage [tex]V[/tex] across this resistor:
[tex]\begin{aligned} V &= I\, R \\ &= 30\; {\rm A} \times 30\; {\rm \Omega} \\ &= 30\; {\rm A} \times 30\; {\rm V \cdot A^{-1}} \\ &= 900\; {\rm V} \end{aligned}[/tex].
The distance between a carbon atom ( m = 12 u ) and an oxygen atom ( m = 16 u ) in the CO molecule is 1.13 × 10-10 m .
How far from the carbon atom is the center of mass of the molecule?
Express your answer to two significant figures and include the appropriate units.
6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom and is the center of mass of the molecule.
The sum of the products of the masses of the two particles and their respective position vectors equals the product of the total mass of the system and the position vector of the center of mass.
The center of mass of CO molecule will be on the line joining C and O atoms. Let the CO molecule be along the X- axis, the C atm being at the origin (x=0). The center of mass relative to the C atom is given by
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
Where, [tex]m_{c}[/tex] and [tex]m_{o}[/tex] are the respective masses of C and O atoms, [tex]x_{c}[/tex] and [tex]x_{o}[/tex] their distances relative to C atom.
Here, [tex]m_{c}[/tex] = 12 u , [tex]m_{o}[/tex] = 16 u and [tex]x_{o} =1.13[/tex]×[tex]10^-^{10}[/tex] m
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
[tex]x_{cm} =[/tex] (12 × 0 + 16 × 1.13 × [tex]10^{-} ^{10}[/tex] ) / (12 + 16 )
[tex]x_{cm}[/tex] = 0.64571 × [tex]10^{-} ^{10}[/tex]
[tex]x_{cm}[/tex] = 6.45 ×[tex]10^{-} ^{11}[/tex] m
Therefore, 6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom is the center of mass of the molecule.
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Have my equation written out but struggling to solve. Can someone help me solve!
The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.
How to find the missing current in a circuit
In this question we must make use of Kirchhoff's laws to find the values of the missing currents in the circuit presented in the picture. There are two rules according to Kirchhoff's laws:
The sum of currents found at nodes of circuits is always equal to zero.The net sum of voltages in a closed loop of a circuit is always equal to zero.Based on the information given by the picture, we have the following system of linear equations that describes the entire circuit:
i₃ = i₁ + i₂
- i₁ · R₁ + ε₁ - i₁ · r₁ - i₁ · R₅ + ε₂ - i₂ · (r₂ + R₂) = 0
ε₂ - i₂ · (r₂ + R₂) - i₃ · r₄ - ε₄ - i₃ · r₃ + ε₃ - i₃ · R₃ = 0
- i₁ - i₂ + i₃ = 0 (1)
(R₁ + r₁ + R₅) · i₁ + (r₂ + R₂) · i₂ = ε₁ + ε₂ (2)
(r₂ + R₂) · i₂ + (r₄ + r₃ + R₃) · i₃ = ε₂ + ε₃ - ε₄ (3)
If we know that R₁ = 5 Ω, r₁ = 0.10 Ω, R₅ = 20 Ω, r₂ = 0.50 Ω, R₂ = 40 Ω, r₄ = 0.20 Ω, r₃ = 0.05 Ω, R₃ = 78 Ω, ε₁ = 22 V, ε₂ = 49 V, ε₃ = 10.5 V and ε₄ = 33 V, then the currents flowing in the circuit are:
- i₁ - i₂ + i₃ = 0
25.1 · i₁ + 40.5 · i₂ = 71
25.1 · i₂ + 78.25 · i₃ = 26.5
The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.
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A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of position and time, y(x,t), is described by the following equation:
[tex]y(x,t)=0.0480[/tex]×[tex]sin(5.40x-128t)[/tex]
where x and y are in meters and the time is in seconds.
1. What is the wavelength of the wave?
2. What is the velocity of the wave? (Define positive velocity along the positive x-axis.)
3. What is the maximum speed in the y-direction of any piece of the string? (Give a positive answer for speed.)
The wavelength of the wave is 1.16m and the velocity is 23.64m/s.
To find the answer, we have to know more about the Transverse waves.
How to find different parameters of a wave?The displacement of the string as a function of position and time, y(x,t), when the wave traveling along a string lying along the x-axis is given as,[tex]y(x,t)=0.048sin(5.40x-128t)[/tex]
Comparing this with the general form of wave equation, we get,[tex]amplitude, a=0.048m\\wave vector, k=5.40\\angular frequency,w=128Hz[/tex]
We have to find the wavelength of the wave, for this, we have the expression as,[tex]k=\frac{2\pi }{wavelength} \\\\wavelength=\frac{2\pi }{k} =\frac{2*3.14}{5.40} \\\\wavelength=1.16m[/tex]
We have to find the velocity of the wave,[tex]v=frequency*wavelength\\2\pi f=w, thus,\\f=\frac{w}{2\pi } =\frac{128}{2*3.14}=20.38s^{-1}\\\\v=1.16*20.38=23.64m/s[/tex]
Thus, we can conclude that, the wavelength of the wave is 1.16m and the velocity is 23.64m/s
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The CM of an empty 1300-kg car is 2.45 m behind the front of the car.
How far from the front of the car will the CM be when two people sit in the front seat 2.70 m from the front of the car, and three people sit in the back seat 3.65 m from the front? Assume that each person has a mass of 60.0 kg .
Express your answer to three significant figures and include the appropriate units.
The center of mass of the car with all people seated will be 3.20 m far from the front of the car.
Taking the front of the car as reference:
m₁x₁ + m₂x₂ + m₃x₃ = mx
where,
m₁ = mass of empty car = 1300 kg
x₁ = distance of center of mass of empty car from front = 2.45 m
m₂ = mass of 2 people sitting in front = 2 x 60 kg = 120 kg
x₂ = distance of center of mass of 2 people sitting in front from the front = 2.7 m
m₃ = mass of 3 people sitting in back seat = 3 x 60 kg = 180 kg
x₃ = distance of center of mass of 3 people sitting in the back seat from the front = 3.65 m
m = total mass of car with all people = 1300 kg + 120 kg + 180 kg = 1600kg
x = distance of center of mass of the car from the front when all are seated
Therefore,
(1300 kg)(2.45 m) + (120 kg)(2.7 m) + (180kg)(3.65 m) = (1300 kg)x
x = 4166 kg.m/1300 kg
x = 3.20 m
Therefore, the center of mass of the car with all people seated will be 3.20 m far from the front of the car.
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three objects are located in a coordinate system as shown below in figure a.find the centre of mass. how does the answer change if the object on the left is displaced upward by 1.00m and the object on the right is displaced downward by 0.500m (b) the object as point particles
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
What is the center of mass of a system of particles?A place at which the entire mass of the body or all the masses of a system of particles appears to be concentrated is known as the center of mass of a body or system of particles. According to physics, the center of mass is a location where the total of the weighted relative positions of the distributed mass's points in space equals zero.a)
The center of mass of a three-particle system is expressed as
[tex]r_{cm}=[/tex] [tex]$$r_{c m}=\frac{\sum_{i=1}^{3} m_{i} r_{i}}{\sum_{i=1}^{3} m_{i}} \Rightarrow \frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$$[/tex]
When the system is only on [tex]$\mathrm{x}$[/tex]-axis (i.e. [tex]$\mathrm{y}=0$[/tex] )
[tex]$$\begin{aligned}x_{c m} &=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\x_{c m} &=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\x_{c m} &=0.5909 \mathrm{~m} \\\text { Therefore } r_{c m}=(0.5909 \mathrm{~m}, 0)\end{aligned}$$[/tex]
b)
When the two particles are shifted
[tex]\begin{aligned}&x_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\&x_{c m}=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\&x_{c m}=0.5909 \mathrm{~m} \\&y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}} \\&y_{c m}=\frac{(5.0 \times 1.0)+(2.0 \times 0)+(4.0 \times 0.5)}{5.0+2.0+4.0} \\&y_{c m}=0.6364 \mathrm{~m}\end{aligned}[/tex]
Therefore [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
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1. If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible for one to be at rest after the collision? Explain.
Answer:
(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?
No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum
(b)Is it possible for only one to be at rest after the collision?
Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.
The graph shows the US Department of Labor noise regulation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a Boom Box at an average level of 86.0 dB.
1. Calculate the INCREASE in the sound level from the ambient work environment level (in dB).
2. A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 116 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?
(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.
(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
Increase in the sound levelThe INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;
Increase in sound level = final sound level - original sound level
Increase in sound level = 86 dB - 85 dB = 1 dB
Factor of sound level increasefrom the graph at 1.5 hours/day, sound level = 97 dB
Increase in sound intensity = final sound level - original sound level
Increase in sound intensity = 116 dB - 97 dB = 19 dB
Factor increase = 19/97 = 0.196 = 19.6%
Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
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*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 330 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
__________m/s²
(b) The electron is initially at rest. What is its speed (in m/s) after 8.00 ✕ 10−⁹ s?
______m/s
The final speed of the electron is 4.64 * 10^5 m/s.
What is the speed of the electron?Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
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a)Calculate the angular velocity of the Earth in its orbit around the Sun.
Express your answer using three significant figures.
b) Calculate the angular velocity of the Earth about its axis.
Express your answer using three significant figures.
The earth's orbital angular velocity around the Sun is 0.986° per day, and its axial angular velocity is 15.041° per hour.
Calculation:
The speeds, not the velocities, are what we require in this case, therefore that is how I will approach it.
Regular speed is calculated by dividing the distance traveled by the travel time.
So, speed = [tex]\frac{distance -travelled}{time}[/tex]
The angular velocity is essentially the same.
Hence, it's divided by the amount of time it took to turn the angle.
(a) Calculation of Earth's orbit around the Sun :
The Earth orbits the Sun in a circle that is 2 radians in diameter (360 degrees). It also takes a year, as we are aware (approx 365 days)
360°/365.25636 days = 0.986°/day ≈ 1° per day
(b) Next, we have to calculate Earth's angular velocity on its axis:
determining the Earth's angular velocity as it fulfills a complete revolution on its axis (a solar day)-
This one requires more precision because a day does not always have 24 hours. Depending on how we define a day,If the day is defined as the time between the Sun's highest and lowest points in the sky, a year's worth of data equals an average of 24 hours.If we define a day as the duration of time it requires for a planet to get to the same location in the sky the following night, then it is 23 hours 56 minutes 4.09 seconds.According to this, the angular speed of rotation = 360°/ 23h 56min 4.09s = 15.041° / hour.
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Please help!! will give brainliest if correct!
B: Scientists assume the universe is a vast single system in which basic laws are consistent.
I ____________ with this statement
Explain your answer, using an example from your investigation about habitable worlds.
Scientists assume the universe is a vast single system in which basic laws are consistent. I agree with this statement
What is the name of the theory that states that our universe is from a single point?The Big Bang Theory is known as one of the greatest consensus that was said to have occurred among scientists, astronomers and also that of cosmologists.
Note that the theory is one that states the Universe as we all know it was formed as a result of a massive explosion that is not only formed by the majority of matter, but also with the use of the physical laws that works in our ever-expanding cosmos.
Therefore, In the case above, i do agree with the fact that Scientists assume the universe is a vast single system in which basic laws that are consistent.
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explain the use of low energy x-ray?
Answer:
The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.
Respond to the following prompts in your initial post:
1. Is Felisha's argument sound or unsound? Use the concepts discussed in your textbook: a valid
structure, true premises, and a sound or unsound argument.
2. Using the fallacies discussed in chapter 11 of your book, identify one fallacy (by name) in
Felisha's argument and explain why this fallacy is an illogical way to argue her conclusion.
3. Describe a time when you successfully argued a point of view and explain why you believe
you were successful.
Based on the textbook, Felisha's argument is known to be unsound because not all the premises that were given was true.
2. She stated that the “The Cowboys are popular and using only their huge fan base to tell is not right and if they are are better as she claim, why are they not the best but she stated that they are among the best.
3. A time when when i successfully argued a point of view was during my high school debate on the topic that doctors are better than lawyers and i gave a lot of valid reasons why the world cannot do without doctors.
What is fallacies and its types?The term fallacies are seen as flawed or deceptive and false kind of arguments that is known to be be proven wrong with the aid of reasoning.
Note that there are two key types of fallacies which are the formal fallacy and an informal fallacy.
Therefore, Based on the textbook, Felisha's argument is seen as an unsound argument because not all the premises that were given was true and they were not standard enough.
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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface. Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?
(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
(b) The work done in firing the projectile is 2,500 J.
Kinetic energy of the projectile at maximum height
The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;
K.E = ¹/₂mv₀ₓ²
where;
m is mass of the projectilev₀ₓ is the initial horizontal component of the velocity at maximum heightNote: At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.
K.E = (0.5)(2)(30²)
K.E = 900 J
Work done in firing the projectileBased on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.
W = K.E(i) = ¹/₂mv²
where;
v is the resultant velocityv = √(30² + 40²)
v = 50 m/s
W = (0.5)(2)(50²)
W = 2,500 J
Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
The work done in firing the projectile is 2,500 J.
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What is meant by significant figures
Answer:
The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.
Answer:
each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first nonzero digit.
Explanation:
A 100 N force pulls a box but it doesn’t move. How much work is done?
A. 0 J
B. 100 J
C. 50 J
D. Not enough info
Answer:
A
Explanation:
Work = force * distance
= 100 n * 0 = 0 work done
why does a bus look red in the daylight?
Answer:
The red color in the sky at sunset (and sunrise) is due to an effect called Rayleigh scattering.
Explanation:
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude and direction of the force on a
−7.35 C charge at this location.
Magnitude _____N
Direction?
- north
-south
-east
-west
Answer:
Approximately [tex]3.86\; {\rm N}[/tex] (given that the magnitude of this charge is [tex]-7.35\; {\rm \mu C}[/tex].)
Explanation:
If a charge of magnitude [tex]q[/tex] is placed in an electric field of magnitude [tex]E[/tex], the magnitude of the electrostatic force on that charge would be [tex]F = E\, q[/tex].
The magnitude of this charge is [tex]q = 7.35\; {\rm \mu C}[/tex]. Apply the unit conversion [tex]1\; {\rm \mu C} = 10^{-6}\; {\rm C}[/tex]:
[tex]\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}[/tex].
An electric field of magnitude [tex]E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}[/tex] would exert on this charge a force with a magnitude of:
[tex]\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}[/tex].
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
An object has a gravitational
potential energy of 750 Joules and is
sitting on a shelf 3.0 meters above
the ground. What is the object's
mass?
[?] kilograms
Answer:
potential energy = (mass × height × acceleration due to gravity.)
750 = mass × 3 × 10
hence, mass = 750/30 = 25 kg.
A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 revolutions before coming to rest.
Find the constant angular acceleration (in rad/s2) of the centrifuge.
______rad/s2
The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
Constant angular acceleration
Apply the following kinematic equation;
ωf² = ωi² - 2αθ
where;
ωf is the final angular velocity when the centrifuge stops = 0ωi is the initial angular velocity θ is angular displacementα is angular accelerationωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s
θ = 52 rev x 2π rad/rev = 326.7 rad
0 = ωi² - 2αθ
α = ωi²/2θ
α = ( 356.05²) / (2 x 326.7)
α = 194.02 rad/s²
Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
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I will run one mile in under 10 minutes 3 months from today.” is an example of a ___________ goal.
Long-term
Short-term
SMART
Both A and B
Answer:
long-term
Explanation:
usually a short term goal is able to be accomplished in a week or two. the question gives a 3 month time frame for the person to build up to the end goal.
Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
A. it gets stuck in the atmosphere
B. it is magnetically attracted
C. it is attached by a rope to the Earth
D. it falls at the same rate the Earth curves
The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.
Newton's Cannonball:Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.
Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.
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PLS HELP 20 PIONTS!!
It is critical for scientists to be able to describe components of a system quantitatively. Explain why it is important to be able to describe a system quantitatively, using an example from your investigation about habitable worlds
A quantitative description of a system is important as it provides information about the carrying capacity of habitable worlds.
What is a quantitative description of the components of a system?Quantitative description refers to the description of a system which is focused on the numerical value of the properties or components of the system.
For example, a quantitative description of the components of a given habitat will be focused on the number of the individual species in the habitat. It will also be focused on the numerical value of the non-living components of the system and how such values affect the living components of the system numerically.
This information will then be used by scientists to see how modifications of the various quantitative components of the habitat will help to improve the chances of survival of species found in the habitat. This leads to such concepts as the carrying capacity of a habitat which is the maximum number of species that the available resources is a habitat can easily sustain.
In conclusion, a quantitative description is important to in investigations about the habitable world.
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formula for conductivity
Answer:
Thermal conductivity formula
k is the thermal conductivity (Wm -1 K -1)Q is the amount of heat transferred through the material (Js -1)A is the area of the body (m 2)ΔT is the temperature difference (K)Explanation:
a dog runs back and forth between its two owners, who are walking toward one another. the dog starts running when the owners are 10 m apart. if the dog runs with a speed of 3 m/s and the owners each walk with a speed of 1.3 m/s how far has the dog traveled when the owners meet
Answer:
11.54 meters
Explanation:
The owners take 10m / (1.3 + 1.3 m/s) = 3.85 seconds to meet
The dog is running at constant speed of 3 m/s for these 3.85 seconds
3 m/s * 3.85 sec = 11.54 meters
a resistor wire of diameter 0.6cm has a resistivity of 1.0×00000010 ohms m. what length of the wire would be needed to make a 4 ohm resistor
The length of the wire is 1.12m
What is Resistance ?
Resistance is opposition to the current flowing in the circuit.
Given ,
D= 0.6m
R= 0.6/2m
R= 28Ω
rho = 1Ωm
R= rho ×l/A
where R is resistance , l is length ,A is area and rho is resistivity
By applying value and Solving we get,
l= 1.12m
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A 29-g rifle bullet traveling 200 m/s embeds itself in a 3.4-kg pendulum hanging on a 2.5-m-long string, which makes the pendulum swing upward in an arc. (Figure 1)
a) Determine the vertical component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.
b)Determine the horizontal component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.
The vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.
What is the velocity of the pendulum just after the collision?As per conversation of momentum, mass of bullet × velocity= mass of pendulum × velocity
=> 0.029kg × 200 = 3.429 × velocity of pendulum
=> Velocity of pendulum= 5.8/3.429 = 1.7 m/s
What is the amplitude of the pendulum's motion?As per conversation of energy,1/2 × mass of pendulum × velocity²=1/2 × k × amplitude²
Here, k = mg/L = (3.429×9.8)/2.5 = 13.44 N/mSo, 3.429×1.7²= 13.44× amplitude²=> 10 = 13.44× amplitude²
=> Amplitude = √(10/13.44)
= 0.86m
What is the angle made by the pendulum with vertical direction at largest displacement?Angle = tan inverse (amplitude/length of pendulum)
= tan inverse (0.86/2.5)
= 20°
What is the maximum displacement along vertical direction?Maximum displacement along vertical direction= L - L×cos20°
= 2.5 - 2.5×cos20°
= 0.15 m
What's the maximum displacement along horizontal direction?Maximum displacement along horizontal direction= Lsin20°
= 2.5 × sin 20°
= 0.85m
Thus, we can conclude that the vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.
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A child in a boat throws a 5.90-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 25.0 kg and the mass of the boat is 38.4 kg. (Figure 1)
Calculate the velocity of the boat immediately after, assuming it was initially at rest.
Express your answer to three significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the box and negative value if the direction of the velocity is in the direction opposite to the velocity of the box.
-0.930 m/s is the velocity of the boat.
Given:Mass of child and boat , [tex]m_1[/tex] = (25.0 + 38.4 )kg
= 63.4 kg
Mass of the package, [tex]m_2[/tex] = 5.90 kg
Velocity of package thrown from boat , [tex]v_2[/tex] = 10.0m/s
[tex]v_1 =?[/tex]
Initial velocity v = 0 m/s
As the boat is at rest, [tex](m_1 + m_2) v=0[/tex]
According, to the law of conversation of momentum;
∴ Momentum before = Momentum after
[tex]( m_1 + m_2 ) v = m_1v_1 + m_2v_2\\0 = m_1v_1 + m_2v_2\\0 = 63.4 v_1 + 5.90*10.0\\63.4 v_1 = - 5.9\\v_1 = - 0.930 m/s[/tex]
Negative direction shows the velocity in the direction opposite to the motion of the package.
Therefore, -0.930 m/s is the velocity of the boat.
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres is 46622.72N
Calculation and Parameters( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.
Pressure difference = (940 - 12)
= 928 millibars.
(928 x 100)
= 92,800N/m^2.
Therefore, the required force to pull the two hemispheres is
(92800 x 0.5024)
= 46622.72N.
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how much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m
2.89watts.
What is meant by sound intensity?The average rate at which sound energy moves across a unit area normal to a given direction is used to determine a sound's intensity. This rate is generally stated in ergs per second per square centimeter.Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:
Formula: [tex]I\frac{P}{4\pi r^{2} }[/tex]
I=1.6x10-3 w/m2
r=12m
[tex]I\frac{P}{4\pi r^{2} }[/tex]
[tex]P=I4\pi r^{2}[/tex]
[tex]=(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )[/tex]
The power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m [tex]=2.89watts.[/tex]
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In transverse waves, there is no _________ motion.
A. Perpendicular
B. Vertical
C. Horizontal
D. Up and down
In a transverse wave, the disturbance are carried forward and the particles are undergoing to and fro motion. Hence in transverse waves, there is no horizontal motion.
A wave can be described as a disturbance that travels through a medium from one location to the other location without transporting any matter.
There are two types of matter,
(i) Transverse wave
(ii) Longitudinal wave
In a transverse wave, the particles of the medium move in a direction perpendicular to the direction of propagation of the wave. That means, the particles undergo to and fro motion about their mean position and moves in the direction of the wave propagation.
Examples are:
Radio wavesLight wavesWater waves when a stone is dropped in the pondWhereas longitudinal waves undergo back and forth motion with respect to the mean position.
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