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Scrat [10]
3 years ago
8

Solve each problem. Be sure to show your work and give a final answer with units rounded to the given number of significant figu

res.
1) Find the voltage on a circuit with a resistance of 12.5 Ω if it has a current of 2.35 A.

2) If a circuit with a 9.0 V battery has a current of 6.2 A, how much resistance is in the circuit?

3) A circuit has three resistors with values of 4.0 Ω, 8.0 Ω, and 12 Ω. What is the equivalent resistance if they are wired in series?

4) A circuit has three resistors with values of 4.0 Ω, 8.0 Ω, and 12 Ω. What is the equivalent resistance if they are wired in parallel?
Physics
1 answer:
Alex Ar [27]3 years ago
8 0

Answer:

1. the voltage will be 2.35×12.5 = 29.4V

2. the resistance would be 9.0/6.2= 1.45ohms

3. in series they will add up thus 4+8+12= 24ohms

4. in parallel it will be 2.18ohms

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A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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