An is a heterogeneous mixture in which visible particles settle

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

vesna_86 [32]

Answer:

$6.63\times 10^8\ N/C$

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

$\dfrac{E}{B}=c\\\\E=Bc$

Put all the values,

$E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C$

So, the magnitude of the electric field is equal to $6.63\times 10^8\ N/C$ .

7 0

2 years ago

How many molecules are there in 39 grams of water

Llana [10]

Answer:

39 g H2O contains 1.3 ×1024molecules H2O.

Explanation:

8 0

3 years ago

Which of the following quantities are unknown?

Sauron [17]

Answer:

Charge of the alpha particle

Explanation:

6 0

3 years ago

Please follow my sis on i.n.s.t.a.g.r.a.m Id - tehzz_13

abruzzese [7]

Answer:

btw But I don't use Insta

4 0

2 years ago

Read 2 more answers

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

2 years ago