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An ac series circuit contains a resistor of 20 ohms, a capacitor of 0.75 microfarads of 120 x 10-3 H. If an effective (rms) volt

pav-90 [236]

Answer:

The effective (rms) current when the circuit is in resonance is 6 A

Explanation:

Given;

resistance of the resistor, R = 20 ohms

capacitance of the capacitor, C = 0.75 microfarads

inductance of the inductor, L = 0.12 H

effective rms voltage, $V_{rms}$ = 120

At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).

The effective (rms) current, = $V_{rms}$ / R

= 120 / 20

= 6 A

Therefore, the effective (rms) current when the circuit is in resonance is 6 A

3 0

3 years ago

A car traveling 77 km/h slows down at a constant 0.48 m/s2 just by "letting up on the gas." A) Calculate the distance the car co

vfiekz [6]

Answer:

(a) 477 m

(b) 44.6 s

(c) 21.16 m

(d) 19.24 m

Explanation:

initial speed, u = 77km/h = 21.4 m/s

acceleration, a = - 0.48 m/s2

final speed v = 0

(a) let the stopping distance is s.

Use third equation of motion

$v^2 = u^2 + 2 a s\\\\0 = 21.4^2 - 2 \times 0.48\times s\\\\s = 477 m$

(b) Let t is time.

Use first equation of motion

v = u + at

0 = 21.4 - 0.48 t

t = 44.6 s

(c) Let the distance is s in first second.

Use second equation of motion

$s = u t + 0.5 at^2\\\\s = 21.4\times 1 - 0.5\times 0.48\times 1\\\\s = 21.4 - 0.24 = 21.16 m$

(d) distance traveled in 5 th second is given by

$s = u + 0.5 a (2 n - 1) \\\\s = 21.4 - 0.5\times 0.48 \times (2\times 5 -1)\\\\s= 21.4 - 2.16 = 19.24 m$

8 0

3 years ago

A circular ride has a radius of 32 m If the time of one revolution of a rider is 0.98 s what is the speed of the rider?

11111nata11111 [884]

Given that,

The radius of a circular path, r = 32 m

The time of one revolution of a rider is 0.98 s.

To find,

The speed of the rider.

Solution,

Let v is the speed of the rider. Speed is equal to total distance divided by time taken.

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 32}{0.98}\\\\v=205.16\ m/s$

So, the speed of the rider is 205.16 m/s.

6 0

3 years ago

In tae-kwon-do, a hand is slammed down onto a target at a speed of 14 m/s and comes to a stop during the 3.0 ms collision. Assum

GrogVix [38]

Answer:

A) The impulse is 11,7 kg.m/s . B) The average force on the hand is 3900N

Explanation:

Givens:

Hand's mass (m)= 0,90kg

Hand's initial velocity (vi)= 14m/s

Time of collision (t)= 3ms

A) Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. We use the formula:

/F/ = m. /Δv/

Where:

F= Force of Impulse

m= Mass

Δv= change in speed ( vf-vi)

Using that formula we get:

/F/= 0,90kg . / (0m/s-14m/s) /

/F/=0,90kg . 14m/s

/F/ = 11,7 kg. m/s

*note that (vf) is 0 because the hand stops in the action, so it's final

velocity = 0

B)The average force is equal to the change in the momentum over the change in time. We use the formula:

/F/= m. (Δv) : Δt

Where:

F= the average force from the target

m= the mass of the hand

Δv= hange in speed (vf/vi)

Δt= change in time

*if we look closely, we find the we alredy have half of the equation because we have alredy calculated m.(Δv) in point A.

We boil the equation down to:

/F/= Impulse : Δt

/F/= 11,7kg.m/s : 0,003s

/F/= 3900N

*we use 0,003s as our time because the given time was 3ms.

*the final result is expressed in Newtons because our final result ends up beeing kg.m/s² = N

8 0

4 years ago

All of the following describe the motion of an object except

Lostsunrise [7]

All describe an objects motion except : mass

8 0

3 years ago