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olchik [2.2K]
3 years ago
5

Which of the following would double the amount of current flowing through a piece of metal wire? Which of the following would do

uble the amount of current flowing through a piece of metal wire? Quarter the voltage across it. Quadruple the voltage across it. Double the voltage across it. Halve the voltage across it.
Physics
1 answer:
Strike441 [17]3 years ago
6 0

Answer:

Double the voltage across it.

Explanation:

From Ohm's Law, we have the relation, as follows:

Voltage = (Current)(Resistance)

If we keep the resistance of the wire as a constant, then we get the relation between voltage and current as follows:

Voltage = (Constant) * Current

Voltage α Current

This shows that the current has a direct proportionality with the voltage. So, if we want to double the current flowing through a piece of metal wire, then we must <u>Double the voltage across it.</u>

It can be confirmed by formula as:

V = IR

Now, doubling the voltage:

V' = 2 IR

substituting value of IR = V:

V' = 2 V

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The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

Explanation:

Given;

mass of block, m =  4 kg

coefficient of kinetic friction, μk = 0.25

angle of inclination, θ = 30°

initial speed of the block, u = 5 m/s

From Newton's second law of motion;

F = ma

a = F/m

Net horizontal force;

∑F = mgsinθ + μkmgcosθ

a = \frac{F_{NET}}{m} = \frac{mgsin \theta + \mu_kmgcos \theta}{m} \\\\a = gsin \theta + \mu_kgcos \theta\\\\a = 9.8sin (30) + 0.25*9.8cos(30)\\\\a = 4.9 + 2.1217\\\\a = 7.022 \ m/s^2

At the  top of the ramp, energy is conserved;

Kinetic energy = potential energy

¹/₂mv² = mgh

¹/₂ v² = gh

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12.5 = 9.8h

h = 12.5/9.8

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Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;

v² = u² + 2ah

v²  = 5² + 2 x 7.022 x 1.28

v²  = 25 + 17.976

v²  = 42.976

v = √42.976

v = 6.56 m/s

Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

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