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n200080 [17]
3 years ago
5

A generator delivers an AC voltage of the form Δv =(82 V) sin (75πt) to a capacitor. Themaximum current in the circuit is 1.00A.

Find the following.
(a) rms voltage of the generator
1 V
(b) frequency of the generator
2 Hz
(c) rms current
3 A
(d) reactance
4 Ω
(e) value of the capacitance
5 F
Physics
1 answer:
Cloud [144]3 years ago
6 0

Answer:

1. 57.99V

2. 37.5Hz

3. 0.7072A

4. 82 ohms

5. 5.18x10^-5F

Explanation:

In answer to this question, we have the Standard equation of AC emf to be

V = V0 x sin ωt

We have

V0 = 82V,

ω = 75π

1.

RMS Voltage =

V0/√2 = 82 /√2

= 82/1.414

= 57.99V

2.

ω = 2π* f

75π = 2πf

Frequency,f = 75π/2π

= 235.5/6.28

= 37.5 Hz

3.

RMS current

= Imax/√2

= 1.00/1.414

= 0.7072A

4.

Reactance

= Vrms/ Irms

= 57.99/0.7072

= 81.999

= 82.0 Ω

5.

Reactance = 1/ ω x C

Reactance = 82

ω = 75π

We put these values into the equation above and make c the subject of the formula

C = 1/82.0 x 75π

C = 1/ 82.0 x 75 x 3.14

C = 1/19311

Capacitance = 5.18x10^-5F

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Katarina [22]

Answer:

V = 15m/s

Explanation:

Given the following data;

Initial velocity = 3m/s

Time = 8secs

Acceleration = 1.5m/s²

To find the final velocity, we would use the first equation of motion;

V = U + at

Substituting into the equation, we have

V = 3 + 1.5*8

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6 0
3 years ago
Eleven wa=eighs 47 kg. her height is 1.63 m. what is her bmi
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Answer:

17.7kg/m^{2}

Explanation:

                       BMI =

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                             = 47kg/(1.63m×1.63m)

                             = 17.689kg/m^{2}

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3 years ago
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tensa zangetsu [6.8K]
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4 years ago
What is the average velocity of a car address for my .7 m to 2 m away from the garage in 3 seconds
stepan [7]

Answer:

v=\frac{5}{3}

Explanation:

So average velocity is calculated as

average velocity=\frac{total displacement}{total time}

Hence in our given case total displacement would be 5 m as the car moves a distance 7m to 2m.

Time taken for this moment is given to be 3 seconds.

Hence substituting back we get the average velocity as

v=\frac{5}{3}

Hence this is the average velocity in this case.

5 0
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While playing catch with my dog I bounced the ball off the ground at a 30-degree angle. It had a range of 6 meters.
baherus [9]

The initial velocity of the ball is 8.2 m/s

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A uniform motion along the horizontal direction, at constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (g=9.8 m/s^2, acceleration of gravity)

The range of a projectile can be derived by the equation of motions along the two directions, and it is found to be:

d=\frac{v^2 sin 2\theta}{g}

where

v is the initial velocity of the projectile

\theta is the angle of projection

g is the acceleration of gravity

For the ball in this problem, we have

\theta=30^{\circ}

d = 6 m is the range

Solving for v, we find the initial velocity:

v=\sqrt{\frac{gd}{sin 2\theta}}=\sqrt{\frac{(9.8)(6)}{sin (2\cdot 30^{\circ})}}=8.2 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
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