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matrenka [14]
3 years ago
14

What must occur for work to be done on an object?

Physics
2 answers:
Slav-nsk [51]3 years ago
6 0

The answer is D, The Object must move.

stiv31 [10]3 years ago
4 0
The answer is D since work done must have displacement..if theres force but no displacement there will be no work done based on W=Fs
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What is mass?Explain
Alexxx [7]

Answer:

The term mass is used to refer to the amount of matter in any given object For instance, a person or object may be weightless on the moon because of the lack of gravity but that same person or object maintains the same mass regardless of location

Explanation:

8 0
3 years ago
A _______ contains several blades of differing thicknesses and is used to measure the clearance between parts.        A. dial in
Likurg_2 [28]
This is D. Feeler Gage
5 0
3 years ago
Explain the increase in pressure of a gas when its volume is decreased at constant<br> temperature.
Dmitriy789 [7]

Answer:

For a gas held at constant temperature, we can apply Boyle's law, which states that the product between the gas pressure and its volume is constant:

PV=const.

where

P is the pressure

V is the volume

As we see from the equation, P and V are inversely proportional to each other: this means that when the volume is decreased, the pressure increases, and vice-versa. The reason for that is that when the volume is decreased, the gas is compressed, so the molecules of the gas come closer to each other, so they collide more frequently with the wall of the container, exerting therefore a greater pressure.

7 0
3 years ago
Canopus, which is in the constellation of Carina and Argo Navis, is 310 ly away. You plan a sight-seeing vacation for Canopus an
vazorg [7]

Answer:

D'=2.933*10^{18}m

Explanation:

From the question we are told that:

Distance D=310ly

Where

1ly=9.46*10^{15}

Generally the equation for  meters will you traverse to reach your vacation destination is mathematically given by

D'=D*1ly

D'=310*9.46*10^{15}

D'=2.933*10^{18}m

6 0
3 years ago
Interactive Solution 9.63 illustrates one way of solving a problem similar to this one. A thin rod has a length of 0.620 m and r
finlep [7]

Answer:

w = 7.89 10⁻² rad/s

Explanation:

We will solve this exercise with the conservation of the annular moment, let's write it in two moments

Initial. With the insect in the center

      L₀ = I w₀

End with the bug on the edge

     L_{f}= I w +  I_{bug} w

The moments of inertia are

For a rod

       I = 1/3 M L²

For the insect, taken as a particle

       I = m L²

The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved

      L₀ =  L_{f}

      I w₀ = I w + I_{bug} w

      w = I / (I +  I_{bug}) w₀

 

      w = I / (I + m L²) w₀

Let's calculate

      w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)²   0.185

      w = 1.43 10⁻³ / 3.352 10³ 0.185

      w = 7.89 10⁻² rad/s

6 0
3 years ago
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