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3 years ago

Which statement describes the role of a resistor in an electrical circuit?

Physics

1 answer:

Arisa [49]3 years ago

6 0

The role of a resistor in an electrical circuit is to hinder the movement of charge through a circuit

Explanation:

A charge is carried through a wire by current (indicated as I in physics S.I connotation). Higher current means the rate at which the charge is flowing through the wire is fast. High resistance of the wire means the charge flows with difficulty hence slower. Resistance is given by the formulae;

R = V/I whereby;

R = resistance

V = voltage

I = current

As you can observe R (resistance) is inversely proportional to I (current). Meaning the higher the resistance the lower the current.

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What of the following is the mathematical form of boyles law?

viva [34]

Answer:

Explanation:

V=1/p

By means of cross multiplication so by that we will have pv=1 which also implies p1v1=p2v2 coz boyles law states that the volume of a given mass of gas is inversely proportional to pressure provided that the temperature in kelvin remains constant

7 0

3 years ago

A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component o

Katen [24]

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, $a=49\ m/s^2$

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

$F=75\ kg\times 49\ m/s^2$

F = 3675 N

Ratio, $R=\dfrac{F}{W}$

$R=\dfrac{3675}{75\times 9.8}=5$

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

$F'=\sqrt{F^2+W^2}$

$F'=\sqrt{(3675)^2+(75\times 9.8)^2}$

F' = 3747.7 N

The direction of force is calculated as :

$\theta=tan^{-1}(\dfrac{W}{F})$

$\theta=tan^{-1}(\dfrac{1}{5})$

$\theta=11.3^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Which is heavier , 1kg of steel or i kg of cork

goblinko [34]

Answer:

Same weight so neither is heavier

6 0

3 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

3 years ago

What is the magnitude of the electric field 17.1 cm directly above an isolated 1.83Ã10â5 C charge?

11111nata11111 [884]

Answer:

Electric field, $E=5.63\times 10^{16}\ N/C$