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3 years ago

What is work - energy theorem ??

Physics

2 answers:

velikii [3]3 years ago

7 0

Answer:

The work-energy theorem explains the idea that the net work - the total work done by all the forces combined - done on an object is equal to the change in the kinetic energy of the object

Explanation:

The work-energy theorem states that the work done by the net force acting on a body is equal to the change produced in the kinetic energy of the body. Let F. be the variable force. We have, W=xf∫xiF.

Elden [556K]3 years ago

3 0

The work-energy theorem explains the idea that the net work - the total work done by all the forces combined - done on an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done.

This idea is expressed in the following equation:

is the total work done

is the change in kinetic energy

is the final kinetic energy

is the initial kinetic energy

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B. It has allowed for faster transmission of Internet signals.

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3 years ago

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The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104

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To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

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Where,

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So replacing we have to

$\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s$

Since $1 Light year = 9.48*10^{15}m$

Then the radius in meters would be

$R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})$

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Then the centripetal acceleration would be

$a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2$

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3 years ago

A volleyball player hit a ball with a mass of 0.25 kg. The average acceleration of the ball was 15.5 m/s². How much force did th

iogann1982 [59]

Force can be obtained by multiplying the mass of the object times the acceleration. Since all of the needed values are given, solving is done as shown below:

Force = mass x acceleration

Force = 0.25 kg x 15.5 m/s^2
Force = 3.875 N

Among the choices, the correct answer is C. 3.87 N

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3 years ago

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melamori03 [73]

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Uranium atoms have 92 electrons and the shell structure is 2.8.

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3 years ago

If the radius of curvature of the bump is 52 m, find the apparent weight of a 62-kg person in your car as you pass over the top

kondaur [170]

Answer:

The answer is below

Explanation:

Driving in your car with a constant speed of 12m/s, you encounter a bump in the road that has a circular cross-section, as indicated in the figure . If the radius of curvature of the bump is 52 m, find the apparent weight of a 62-kg person in your car as you pass over the top of the bump.

Solution:

Centripetal force is the net force acting on a body which makes it move along a curved path. This force is always towards the center of curvature.

As the car passes over the bump, the centripetal acceleration acts downward towards the circle center.

The sum of all vertical forces is equal to zero, hence:

F - mg + ma = 0

where F is the apparent weight of the person, m is the mass of the person, ma = centripetal force = mv²/r

Given that:

m = 62 kg, v = velocity = 12 m/s, r = radius of curvature of bump = 52 m, g = acceleration due to gravity = 10 m/s. Therefore:

F - mg + ma = 0

F - mg + mv²/r = 0

F = mg - mv²/r

F = m(g - v²/r)

Substituting:

F = 62(10 - 12²/54)

F = 456.67 N

The apparent weight of a 62-kg person as the top of the bump is passed = 456.67 N

But the weight of the person = mg = 62* 10 = 620 N

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3 years ago

What is work - energy theorem ??​

What is work - energy theorem ??