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tino4ka555 [31]
3 years ago
10

Can anyone help please ​

Physics
1 answer:
Radda [10]3 years ago
4 0

Answer:

Option C, the current is constant through the wire.

Explanation:

Ok, so we know that as larger cross-section we have, smaller resistance, so makes sense to think that the current will be larger in the first piece of wire.

But now let's think in a simpler case.

Suppose that in the image we see pipes, and through these pipes, there is water flowing.

You would say that the water flow in WX is larger than the current in XY?

No, it can´t be, the volume of water that leaves WX in a lapse of time Δt, must be the exact same volume of water that enters XY in that lapse of time, then the water flow is constant.

The only thing that can happen is that the velocity at which the water moves in WX is smaller than the velocity at which the water moves in the other parts.

The same thing happens here, the current will be constant, and the thing that changes is the velocity at which the electrons flow in each part of the wire.

So if you actually had a wire like this, you would see that the thinner parts heat a lot more than the others, and this happens because the electrons in those parts have a lot more kinetic energy.

The correct graph is C.

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Alexeev081 [22]

Answer:

a) c=1822.3214\ J.kg^{-1}.K^{-1}

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

  • mass of aluminium, m_a=0.1\ kg
  • mass of water, m_w=0.25\ kg
  • initial temperature of the system, T_i=10^{\circ}C
  • mass of copper block, m_c=0.1\ kg
  • temperature of copper block, T_c=50^{\circ}C
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  • final equilibrium temperature, T_f=20^{\circ}C

We have,

specific heat of aluminium, c_a=910\ J.kg^{-1}.K^{-1}

specific heat of copper, c_c=390\ J.kg^{-1}.K^{-1}

specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)

Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)

Q_{in}=11375\ J

The heat released by the blocks when dipped into water:

Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)

where

c= specific heat of the unknown material

For the conservation of energy : Q_{in}=Q_{out}

so,

11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)

c=1822.3214\ J.kg^{-1}.K^{-1}

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

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