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3 years ago

The linear expansion of a material depends on which of the following?

Physics

1 answer:

nordsb [41]3 years ago

7 0

Answer:

1. The change in temperature of the material.

2. The length of the material.

3. The type of material.

Explanation:

The term linear expansion is defined as the increase in length of any conductor when there is a change in temperature. The new length of any wire or rod is given by :

$L=L_o(1+\alpha \Delta T)$

$L_o$ is the initial length of the rod

$\alpha$ is the coefficient of linear expansion

$\Delta T$ is the change in temperature

It is clear that the linear expansion of a material depends on :

1. The change in temperature of the material.

2. The length of the material.

3. The type of material.

So, all options are correct.

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A wheel of mass 60.0 kg is pushed with a net force of 26.4 N. Find acceleration.

worty [1.4K]

F= ma; a= F/m

a = 26.4 N/60 kg= 0.44 m/s^2

4 0

3 years ago

A soccer ball kicked with a force of 12.5 N accelerates at 6.2 m/s’ to the right. What is the mass of the ball? Answer in units

stepladder [879]

Answer:

m = 2.01[kg]

Explanation:

This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

$F =m*a$

where:

F = force = 12.5 [N]

m = mass [kg]

a = acceleration = 6.2 [m/s²]

$12.5=m*6.2\\m = 2.01[kg]$

3 0

3 years ago

A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime

Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

7 0

2 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

4 0

3 years ago

IF you are in Space and push a bowling what happens to you and the bowling ball?

Anastaziya [24]

Answer:

The bowling ball did not change size or shape- the only thing that changed was the amount of gravity that pulls on it. But the mass of the bowling ball would never change. A bowling ball with a mass of 12 pounds on earth will have the mass of 12 pounds on the moon! Mass is the amount of atoms that a space fills.

Explanation:

I hope this helps! :D

4 0

3 years ago