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2 years 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 charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

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#LearnwithBrainly

5 0

3 years ago

In general, metalloids are slightly reactive.

algol [13]

Yes, in general, metalloids are slightly reactive.

6 0

4 years ago

Read 2 more answers

If velocity is decreasing,then acceleration:

ziro4ka [17]

If velocity is decreasing, then acceleration is in the direction
opposite to the velocity.

If the object is moving in the direction that you call 'positive',
then acceleration is negative.

5 0

3 years ago

The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of

jasenka [17]

1) Frequency: $3.29\cdot 10^{15}Hz$

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

$E=2.18 aJ=2.18\cdot 10^{-18} J$

The energy of a photon is given by

$E=hf$

where $h=6.63\cdot 10^{-34}Js$ is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

$f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz$

2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

$\lambda=\frac{c}{f}$

where $c=3\cdot 10^8 m/s$ is the speed of light and f is the frequency. Substituting the frequency, we find

$\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm$

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3 years ago

If a sample of an unknown material with a mass of 0.68 g and a volume of 0.8 cm3 is dropped

Anna71 [15]

As the density of the unknown substance is 0.68g/0.8ml = 0.85g/ml, it is less dense than the maple syrup at 1.33g/ml and will float.

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3 years ago