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2 years ago

When we add or remove energy from a substance, what kind of changes can we observe? Can they happen at the same time ?

Physics

1 answer:

const2013 [10]2 years ago

7 0

Answer:

When you add energy to a substance?

One change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules. The temperature at which a substance goes from a solid to a liquid is it melting point.

When you remove energy from a substance?

When a substance is heated, it gains thermal energy. Therefore, its particles move faster and its temperature rises. When a substance is cooled, it loses thermal energy, which causes its particles to move more slowly and its temperature to drop.

Explanation:

The addition of energy increases the kinetic energy of the particles, which reduces the intermolecular forces between the particles. Freezing occurs when a liquid becomes a solid and energy is released.

Atoms lose energy when they change from solid to liquid or gas and quid to gas.

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A proton is placed in an electric field of intensity 700 n/

kakasveta [241]

F=ma
F=QE = 1.602e-19C*700N/C = 1.1214e-16N
1.1214e-16N = ma = 1.6726e-27kg * a
a = 6.702e10 m/s² along the direction of the field line

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3 years ago

If time travel to the future existed, doesn't that mean that the future has already occurred and that we are living in the past?

kodGreya [7K]

Answer:

As of right now the techology has not been invented to time travel

if we were to time travel to the future from where that person travled from would be the past and to them the people from where they came from are living in the past

Explanation:

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3 years ago

If you're walking on the ice cream at 5 ounces per toaster, and your bicycle loses a sock, how much gravy will you need to repai

Lesechka [4]

Answer:

False you dont repaint your hamster.

Explanation:

LOL

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3 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

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3 years ago

A bowling ball with a mass of 7.10 kg has a velocity of 5 m/s. what is the bowling ball's momentum?

melomori [17]

P = mv

P momentum
m mass
v velocity

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3 years ago