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Dahasolnce [82]
4 years ago
7

A resistor, an inductor, and ideal battery, and a switch are connected in series. Just after the switch is closed, which of the

following are zero?
a) the current and the voltage across the inductor
b) the current and the voltage across the resistor
c) the voltage across the inductor and the voltage across the resistor
d) all of the above.
Explain your reasoning.
Physics
1 answer:
Marianna [84]4 years ago
5 0

Answer:

b) the current and the voltage across the resistor

Explanation:

As soon as we close the switch then the current in the circuit will be zero because the inductor connected in series will not allow to change the current through it due to its inertial property

Since initial current through the inductor is zero so after connecting the key the current will still remains zero and then it will start increasing

The equation for current is given as

i = i_o(1 - e^{-Rt/L})

so as soon as the key is closed the current in the circuit is zero and hence the voltage across the resistor is also zero as we know

V_r = i R = 0

so correct answer will be

b) the current and the voltage across the resistor

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Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

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Answer:

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Explanation:

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Answer:

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Explanation:

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