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nevsk [136]
3 years ago
6

A 20.00-ohm, 5.00-watt resistor is placed in series with a power supply.(a) What is the maximum voltage that can be applied to t

he resistor without harmingthe resistor?(b) What would be the current through the resistor?
Physics
1 answer:
Nutka1998 [239]3 years ago
7 0
<h2>Answer:</h2>

(a) 10.00V

(b) 0.5V

<h2>Explanation:</h2>

(a) The power(P) supplied by a resistor of resistance (R) when a voltage(V) is passed across is given by;

P = V² / R       ------------------(i)

From the question;

P = 5.00W

R = 20.00Ω

Substitute these values into equation (i) as follows;

5.00 = V² / 20.00

V² = 5.00 x 20.00

V² = 100.00

V = \sqrt{100.00}

Solve for V;

V = 10.00V

Therefore, the maximum voltage that can be applied without harming the resistor is 10.00V

(b) By Ohm's law the current (I) flowing through a resistor of resistance (R) when a voltage (V) is applied is given by;

V = I x R

=> I = V / R             ----------------(ii)

Substitute the values of V and R into equation (ii) as follows;

I = 10.00 / 20.00

I = 0.5A

Therefore, the current through the resistor is 0.5A

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5 0
2 years ago
Over a period of operation, the useful work output of the fluorescent bulb was
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Answer:

199.0521 Will be the answer

5 0
2 years ago
Which of the following statements does NOT accurately describe the Berlin Airlift?
Karo-lina-s [1.5K]

The Berlin Airlift is best described as the aircraft used to delivered needed food and supplies to the city of Berlin through the air because all other routes were blocked by the Soviet Union.

<h3>What is Berlin Airlift?</h3>

The Berlin airlift was a 1940s military operation that supplied West Berlin with food and other vital goods by air after the Soviet Union blockaded the city.

Thus, the Berlin Airlift is best described as the aircraft used to delivered needed food and supplies to the city of Berlin through the air because all other routes were blocked by the Soviet Union.

Learn more about Berlin Airlift here: brainly.com/question/1104371

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8 0
2 years ago
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tensio
BARSIC [14]

Answer:

C. At the bottom of the circle.

Explanation:

Lets take

Radius of the circle = r

Mass = m

Tension = T

Angular speed = ω

The radial acceleration towards = a

a= ω² r

Weight due to gravity = mg

<h3>At the bottom condition</h3>

T - m g = m a

T =  m ω² r  + m g

<h3>At the top condition</h3>

T + m g = m a

T=  m ω² r -m g

From above equation we can say that tension is grater when ball at bottom of the vertical circle.

Therefore the answer is C.

C. At the bottom of the circle.

8 0
3 years ago
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