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11111nata11111 [884]
3 years ago
10

A capacitor is connected to an ac power supply. The maximum current is 4.0 A and the reactance of the capacitor is Xc = 8.0 Ω. W

hat is the average power dissipated by the capacitor? Select one: a. 32 W b.0 W c. 128 W d. 64 W
Physics
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

option (b)

Explanation:

Io = 4 A

Xc = 8 ohm

The formula for the average power dissipated is given by

P = Vrms x Irms x cos Ф

Where, Ф be the phase difference between Vrms and Irms .

As we know that in case of capacitor, the current leads the voltage by 90 degree, it means the value of Ф is 90 degree.

Thus, Average power = Vrms x Irms x Cos 90

Average power = 0 Watt

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A charged particle moves through a magnetic field. In which situation is the magnetic force zero?
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Answer:

The answer is the option a.

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Fm = q (v x B)

Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.

"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is

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2 years ago
The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti
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Answer: The answer is B

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2 years ago
Read 2 more answers
A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak
GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg          

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.  

From the attached figure, the x and y component of forces is given by :

T_{1x} =-T_1 cos (20)

T_{2x} = T_2 cos (40)

mg_x = 0

T_{1y} = T_1 sin (20)

T_{2y} = T_2 sin (40)

mg_y= -mg

Let R_x and R_y is the resultant in x and y direction.

R_x=-T_1 cos (20)+T_2 cos (40)+0

R_y=T_1 sin(20)+T_2 sin(40)-mg

As the system is balanced the net force acting on it is 0. So,

-T_1 cos (20)+T_2 cos (40)+0=0.............(1)

T_1 sin(20)+T_2 sin(40)-100\times 9.8=0..................(2)

On solving equation (1) and (2) we get:  

T_1=866.86\ N (tension on the left rope)

T_2=1063.36\ N (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.                            

7 0
3 years ago
The power of the kettle was 2.6 kW
faltersainse [42]

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f  = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

7 0
2 years ago
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