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Ostrovityanka [42]
3 years ago
5

A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr

ent leads the emf of the generator by 0.850 rad. A) What is the impedance of the circuit? B) What is the resistance of the circuit? C) Is the circuit predominantly inductive, or capacitive?
Physics
1 answer:
weqwewe [10]3 years ago
3 0

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

\phi = 0.85 rad = 48.72°

(a). Impedance of the circuit is given by

Z = \frac{V}{I}

Z = \frac{150}{2.75}

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }

Put the values of Z & \phi in above formula we get

R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

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a ball rolls off the edge of a 2m high shelf at a speed of 5 m/s and hits the ground the taken to hit the ground is
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Answer:

Explanation:

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v_0=5, and

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When you factor this however your math/physics teacher has you factoring you will get that

t = 1.3 sec and t = -.31 sec

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3 0
2 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
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r is the distance measured from the centre of the planet.

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r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

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4 0
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Answer:

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\omega = 25.13 rad/s

5 0
3 years ago
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