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2 years ago

8

Distance (m)

1 answer:

vitfil [10]2 years ago

3 0

Answer:41 m/s

Explanation:

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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

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Practice 3: Label the correct phase that would result from the Moon and Earth in these positions.

Anna71 [15]

Answer:

both position I think in nor

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How many neutrons are in an atom of Carbon-13 or (613C)?

vivado [14]

Answer: 6 OR 7

Explanation:

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2 years ago

Plz i need help fast im timed!!!!!!

professor190 [17]

Answer:

sorry i throght i had the answer

Explanation:

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1 year ago

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A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

so final speed will be 2.41 rad/s

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2 years ago