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1 year ago

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

1 answer:

7 0

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

t=0.476v
t=1.967v
V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

t = 1.0 seconds
5.0 seconds
20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

t=0.476v
t=1.967v
V2=4.323v

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A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

2 years ago

If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t

Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

μ₀ =permeability of free space 4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0

2 years ago

A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times be

ra1l [238]

Answer:

Explanation:

Given,

initial angular speed, ω = 3,700 rev/min

= $3700\times \dfrac{2\pi}{60}=387.27\ rad/s$

final angular speed = 0 rad/s

Number of time it rotates= 46 times

angular displacement, θ = 2π x 46 = 92 π

Angular acceleration

$\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}$

$\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}$

$\alpha = -259.28 rad/s^2$

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2 years ago

When a deck of cards slips out of your hands and falls to the ground, gravity<br> does work on it

motikmotik

Answer:

yes

Explanation:

3 0

2 years ago

Read 2 more answers

A series RLC circuit has a resonant frequency = 6.00 kHz. When it is driven at a frequency = 8.00 kHz, it has an

ANEK [815]

The resistance (R) of the circuit is 707.1 ohms and the inductance (L) is 0.032 H.

<h3>Resistance of the circuit</h3>

For the phase constant of 45⁰, impedance is equal to the resistance of the circuit.

$Z= R\sqrt{2} \\\\R = \frac{Z}{\sqrt{2} } \\\\R = \frac{1000}{\sqrt{2} } = 707.1 \ ohms$

<h3>Resonant frequency</h3>

$f = \frac{1}{2\pi \sqrt{LC} } \\\\6000 = \frac{1}{2\pi \sqrt{LC} } \\\\2\pi(6000) = \frac{1}{\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi (6000)} \\\\LC = (\frac{1}{2\pi (6000)})^2\\\\LC = 7.034 \times 10^{-10} \\\\ C = \frac{7.034 \times 10^{-10} }{L} ---(1)$

<h3>At driven frequency</h3>

$X_l- X_c = R\\\\\omega L - \frac{1}{\omega C} = 707.1\\\\2\pi f L - \frac{1}{2\pi fC} = 707.1\\\\2\pi (8000) L - \frac{1}{2\pi (8000) C } = 707.1\ \ --(2)\\\\$

<em>solve 1 and 2 together</em>

$2\pi(8000) L - \frac{L}{2\pi (8000)(7.034 \times 10^{-10})} = 707.1\\\\50272L - 28279.48L = 707.1\\\\L = 0.032 \ H$

Learn more about impedance of RLC circuit here: brainly.com/question/372577

7 0

2 years ago