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sweet [91]
3 years ago
14

Consider two wires made of copper. They have the same diameter but one of them has three times the length of the other. Choose t

he correct statement:
a. Both wires have the same resistivity
b. The long wire has one third of the resistivity of the short wire
c. None of the other answers is correct
d. The long wire has three times the resistivity of the short wire
Physics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

a. Both wires have the same resistivity

Explanation:

For the resistance of a wire , following formula holds good .

R = ρ l / S , R is resistance , l is length , S is cross sectional area and ρ is resistivity of the material that the wire is made of. Resistance is dependent on length and cross sectional area but resistivity does not depend upon length or cross sectional area . It only depends upon the type of material.

If we replace  copper wire with aluminium wire , then resistivity will change .

Hence , since the wire remains made of copper , resistivity will not change.

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A rocket explodes into two fragments, one 25 times heavier than the other. The magnitude of the momentum change of the lighter f
V125BC [204]

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0  = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

<u>So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.  </u>

So, option B is the answer

4 0
3 years ago
Give examples of friction being useful and not useful in everyday life
Aloiza [94]
If you are cold and you rub your hands together then that friction can create warmth. Useful

If you're sliding down a staircase railing and your hands are on the raiking then the friction may burn your hands a little. Not useful<span />
6 0
3 years ago
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Which statements explain what happens at deep-ocean trenches? Check all that apply.
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Answer:

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An object pulled to the right by two forces has an acceleration of 2.5m/s2. The free-body diagram shows the forces acting on the
Vanyuwa [196]

Answer:

490 N is the correct answer.

Explanation:

8 0
3 years ago
Help with physical science please
alex41 [277]

1. Elastic potential energy (D. EEl)

In this situation, the spring is compressed with the toy on top of it. The toy is stationary, so it does not have kinetic energy. However, the spring is compressed, so it does have elastic potential energy, given by:

E_{EL}=\frac{1}{2}kx^2

where k is the spring constant and x is the compression of the spring.

2. Gravitational potential energy (C. Eg)

In this situation, the spring has been released, so it returns to its natural position, so its elastic potential energy is zero. The toy is also stationary, since it is at its top position, where its velocity is zero, so its kinetic energy is also zero. However, the toy is now at a certain height h above the spring, so it has gravitational potential energy given by:

E_g = mgh

where m is the mass of the toy and g is the gravitational acceleration.

3. Gravitational potential and kinetic energy (A. Eg and EK)

In this situation, the toy is falling: so, it is moving with a certain speed v, so it has kinetic energy given by

E_k = \frac{1}{2}mv^2

Also, since it is at a certain height above the spring, it still has some gravitational potential energy, as in the previous point.

4. Gravitational potential energy (C. Eg)

The jumper is standing on the bridge, so it has gravitational potential energy given by its height h above the ground:

E_g=mgh

where m is the mass of the jumper.

5. This exercise has the same text of the previous one.

8 0
3 years ago
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