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3 years ago

7

A positive charge of 6.0 x 10-4 C is in an electric field that exerts a force of 4.5 x 10 -4 on it. What is the strength of the

electric field?

1 answer:

Yuki888 [10]3 years ago

6 0

Electric field is defined as force per unit charge.

So it is given by

$F = q E$

now we can find electric field by

$E = \frac{F}{q}$

$E = \frac{4.5*10^{-4}}{6 * 10^{-4}}$

$E = 0.75 N/C$

So field strength is 0.75 N/C.

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3 years ago

Một chất điểm chuyển động tròn đều trên một cung tròn bán kính 40 cm. Biết chất điểm đi được 5 vòng trong thời gian là 2 s. Tính

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3 years ago

A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration

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3 years ago

a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec

MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

$\frac{1}{2} mu_1^2+0=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

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It is impossible for the mass to be zero.
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An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

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3 0

1 year ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago