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Gnom [1K]
4 years ago
13

You push on a 4 N object for 10 meters. find work

Physics
2 answers:
alekssr [168]4 years ago
5 0

Answer:

40 J

Explanation:

Work= ForcexDistance

Gnom [1K]4 years ago
4 0

Answer:

40J

Explanation:

work is force x distance = 4 x 10 = 40 J

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There is a 90 kg woman attached at the end of a bungee cord (k = 35 N/m) that is experiencing simple harmonic motion. How long d
Cloud [144]

Answer:

The value is T  =10.1 \ s

Explanation:

From the question we are told that

    The mass of the woman is  m  = 90 \  kg  

    The spring constant of the bungee cord is  k  =  35 \  N/  m

Generally the period of the oscillation (i,e time taken to complete on  cycle ) is mathematically represented as

            T  = 2 \pi *  \sqrt{ \frac{m}{k} }

=>      T  = 2 * 3.142  *  \sqrt{ \frac{90 }{ 35} }

=>      T  =10.1 \ s

3 0
3 years ago
A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box
Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the  force is parallel to the displacement, \theta=0

Solving the equation for F, we find the force exerted on the box:

F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
3 years ago
Your friend says that for a moving object to continue moving, a force must be continually applied to it. Do you agree with your
Zigmanuir [339]

Answer:

I would disagree with my friend because that according to Newton's first law, an object in motion will continue to be in motion until stopped by another object/force. If the object is already moving, it will stay in motion until something else stops it. There is no need for a force to be <em>continually</em> applied to it while it's already moving.

4 0
2 years ago
Help with physical science please
alex41 [277]

1. Elastic potential energy (D. EEl)

In this situation, the spring is compressed with the toy on top of it. The toy is stationary, so it does not have kinetic energy. However, the spring is compressed, so it does have elastic potential energy, given by:

E_{EL}=\frac{1}{2}kx^2

where k is the spring constant and x is the compression of the spring.

2. Gravitational potential energy (C. Eg)

In this situation, the spring has been released, so it returns to its natural position, so its elastic potential energy is zero. The toy is also stationary, since it is at its top position, where its velocity is zero, so its kinetic energy is also zero. However, the toy is now at a certain height h above the spring, so it has gravitational potential energy given by:

E_g = mgh

where m is the mass of the toy and g is the gravitational acceleration.

3. Gravitational potential and kinetic energy (A. Eg and EK)

In this situation, the toy is falling: so, it is moving with a certain speed v, so it has kinetic energy given by

E_k = \frac{1}{2}mv^2

Also, since it is at a certain height above the spring, it still has some gravitational potential energy, as in the previous point.

4. Gravitational potential energy (C. Eg)

The jumper is standing on the bridge, so it has gravitational potential energy given by its height h above the ground:

E_g=mgh

where m is the mass of the jumper.

5. This exercise has the same text of the previous one.

8 0
3 years ago
A very small source of light that radiates uniformly in all directions produces an electric field amplitude of 8.45 V/m at a poi
Mandarinka [93]

Answer:

4.43 kW

Explanation:

Since Intensity I = P/A = E²/2cμ₀ where P = Power, A = Area = 4πr² where r = distance from source = 61 m and E = electric field amplitude = 8.45 V/m.

P = E²A/2cμ₀ = E²4πr²/2cμ₀ = 2πE²r²/cμ₀

  = 2π(8.45 V/m)²(61 m)²/3 × 10⁸ m/s × 4π × 10⁻⁷ Tm/A

  = 4428.1 W

  = 4.4281 kW ≅ 4.43 kW

6 0
3 years ago
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