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Leto [7]
3 years ago
5

If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil

l
A decrease.
B increase.
C increase, then decrease.
D not change.
Physics
2 answers:
mina [271]3 years ago
6 0
The answer is A.) decrease.
Take pushing a box for example. You push your hardest but then give out, still trying to push. You're doing less work than what you started with.
Dominik [7]3 years ago
6 0

Answer:

Option A is the correct answer.

Explanation:

We know that the expression for work

          Work = Force x Displacement

So work is directly proportional to force.

Here force is decreasing, so work also decreases.

Option A is the correct answer.

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I WILL GIVE U BRAINLIST layers: Earth:: ___: the Sun
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Answer:

There are layers on the Earth.  There is gas on the Sun.

This means that the answer is...

D: Gas

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3 years ago
The displacement vector from your house to the library is 520 m long, pointing 40 ∘ north of east.What are the x-component (x-ax
Alexeev081 [22]

Answer:

398.3 m, 334.2 m

Explanation:

The magnitude of the displacement vector is

v = 520 m

And its direction is

\theta=40^{\circ}

measured as north of east.

The x-component of this vector is given by:

v_x = v_0 cos \theta = (520)cos 40^{\circ}=398.3 m

While the y-component is given by

v_y = v_0 sin \theta =(520)sin 40^{\circ}=334.2 m

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Momentum quiz for physics
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Answer:

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Use the drop-down menu to answer the question. "Take a Closer Look" shows the Hubble telescope. Where is the Hubble telescope lo
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5 0
2 years ago
Read 2 more answers
A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ s
irina1246 [14]

Answer:

v_{f}\approx 2.097\,\frac{m}{s}

Explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

m_{cyl}\cdot g\cdot y_{o} = \frac{1}{2}\cdot m_{cyl} \left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

g\cdot y_{o} = \frac{1}{2}\cdot\left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

The translational speed is:

v_{f} = \sqrt{\frac{2\cdot g\cdot y_{o}}{\left(1 + \frac{1}{R}  \right)}}

v_{f} = \sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (4.80\,m)}{\left(1 + \frac{1}{0.049\,m}  \right)} }

v_{f}\approx 2.097\,\frac{m}{s}

7 0
3 years ago
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