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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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Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
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Answer:

W=2.76\times 10^{-6}\ J

Explanation:

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Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

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Answer:

b. craters, river valleys feeding into surface lakes of very cold liquids

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A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
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Answer:

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  We have force applied is also equal to product of mass and acceleration.

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