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3 years ago

5

An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then d

isconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

1 answer:

Basile [38]3 years ago

7 0

Answer:

$p.d' = 37.6 V$

Explanation:

From the question we are told that:

Potential difference $p.d=18.8V$

New Capacitor $C_1=C_2/2$

Generally the equation for Capacitor capacitance is mathematically given by

$C=\frac{eA}{d}$

Generally the equation for New p.d' is mathematically given by

$C_2V=C_1*p.d'$

$p.d' = 2V$

$p.d'= 2 * 18.8$

$p.d' = 37.6 V$

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two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second

Fittoniya [83]

Answer:

Approximately $0.88\; {\rm m \cdot s^{-1}}$ to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Since momentum of this system (of the astronauts) conserved:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}$ .

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be $0$ since the velocity of both astronauts was $0\!$ .

Therefore:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

The final momentum of the first astronaut ( $m = 64\; {\rm kg}$ , $v = 0.8\; {\rm m\cdot s^{-1}}$ to the left) would be $p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}}$ to the left.

Let $p_{2}$ denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be $(p_{1} + p_{2})$ .

$\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

Hence, $p_{2} = (-p_{1})$ . In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude ( $51.2\; {\rm kg \cdot m \cdot s^{-1}}$ ) but opposite in direction (to the right versus to the left.)

Rearrange the equation $p = m\, v$ to obtain an expression for velocity in terms of momentum and mass: $v = (p / m)$ .

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}$ .

Hence, the velocity of the astronaut in question ( $m = 58.2\; {\rm kg}$ ) would be $0.88\; {\rm m \cdot s^{-1}}$ to the right.

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2 years ago

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

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3 years ago

What is the momentum of a 65kg ball rolling at 5 m/s?

mojhsa [17]

Explanation:

Momentum = mass × velocity

p = (65 kg) (5 m/s)

p = 325 kg m/s

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3 years ago

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astra-53 [7]

<h2>Answer</h2>

The volume will be <u>increased</u>

<h2>Explanation</h2>

Look at the formula

$Density = Mass/Volume$

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aleksklad [387]

<span>Which of the following is the correct unit for time when calculating power in watts?
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The correct unit for time when calculating power in watts is seconds. The answer is letter D. The rest of the choices do not answer the question above.

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