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3 years ago

True or False: Cooler objects always have more thermal energy than warmer objects.

Physics

1 answer:

tensa zangetsu [6.8K]3 years ago

3 0

False

Because you can see using a thermal camera I guess

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You push a refrigerator with a force of 100 N. If you move the refrigerator a distance of 5 m while you are pushing, how much wo

ddd [48]

The work done to push the refrigerator is 500 Nm.

Explanation:

Work done is the measure of force required to move any object from one point to another. So it is calculated as the product of force and displacement.

If the force increases the work done will increase and similarly, the increase in displacement increases the work done. So to push the refrigerator work should be done on the object and not by the object.

As the force is 100 N and the displacement is 5 m then, work done can be measured as

Work = Force × Displacement

Work = 100 × 5 = 500 Nm

So the work done to push the refrigerator is 500 Nm.

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3 years ago

What do the vertical columns in the periodic table indicate?

romanna [79]

The correct answer is

D. Groups and Families

I did the quiz nd this was the right answer

Hopes this helps :)

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3 years ago

Read 2 more answers

Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide deta

Paladinen [302]

Answer:

She should explain that the Sun is made up of gaseous layers that surround an iron core.

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3 years ago

Help, please<br> and explain

11Alexandr11 [23.1K]

We could tell a force is acting on an object if the object is being pushed, pulled, or moved in any way

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3 years ago

Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl

UkoKoshka [18]

Answer:

a) $W_g =61.25J$

b) $W_k = -46.25J$

c) $W_N = 0$

d) $W_g$ would be the same.

$W_k$ would decrease.

$W_N$ would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

$F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J$

b) The friction force is equal to the normal force times coefficient of friction.

$F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J$

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

$h = x\sin(\theta)$

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of $x\sin(\theta)$ .

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0

3 years ago