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WITCHER [35]
3 years ago
14

An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic

field B = 0.4 T at a constant frequency of 55 Hz. What is the maximum induced emf?
a. 625 V
b. 110 V
c. 421 V
d. 332 V
e. 200 V
Physics
1 answer:
mezya [45]3 years ago
7 0

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

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Answer:

1.56 × 10^-3 cm.

Explanation:

So, we are given the following parameters from the question above;

Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.

Step one: find the area of the computer chip.

Therefore, Area = Length × breadth.

Area = 3.67 cm × 2.93 cm.

Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.

Step two: find the area of one transistor

The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).

Hence;

The area of one transistor= 10.7531/4.4 × 10^6.

The area of one transistor= 2.44 × 10^-6 cm^2.

=> Note that We have our transistors as square, therefore;

The maximum dimension = √ (2.44 × 10^-6) cm^2.

The maximum dimension= 1.56 × 10^-3 cm.

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3 years ago
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tankabanditka [31]
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.

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Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
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suppose you are walking home after school. The distance from school to your home is five kilometers. on foot you can get home in
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3 years ago
Read 2 more answers
You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
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<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
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Answer

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Explanation:

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