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erik [133]
3 years ago
9

A 24 V battery is connected in series to a resistor of 4 Ω, a resistor of 12 Ω and a resistor of 15 Ω. What is the current in th

e 12 Ω resistor?
A. 0.77 A
B. 1.6 A
C. 2 A
D. 6 A
Physics
1 answer:
JulsSmile [24]3 years ago
5 0
Resistance=12Ω
Voltage=24v
Current=voltage/resistance
Current=24/12= 2A
Answer is option C
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A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

<u>Average acceleration</u>

F=ma, so a=\frac{F}{m_{b}}.

Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
What must be the acceleration of a train in order for it to stop 12 m/s in a distance if 541 m
earnstyle [38]

Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

Lets explain how to solve the problem

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

We want to find its acceleration

The acceleration will be negative quantity because the train reduced its

velocity from 12 m/s to zero

We need rule contains velocity, acceleration and distance

So we will use ⇒<em> v² = u² + 2as</em>, where v is the final velocity, u is the

initial velocity, a is the acceleration and s is the distance

v = 0, u = 12 m/s, s = 541 m

Substitute these values in the rule

(0)² = (12)² + 2(a)(541)

0 = 144 + 1082 a

Subtract 144 from both sides

-144 = 1082 a

Divide both sides by 1082

- 0.133 = a

<em>The acceleration of the train must be - 0.133 m/s²</em>

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3 years ago
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irina1246 [14]

Answer:

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In order to find the average velocity of the car we need to know the final and initial positions, and the time that took to get from one to the other.

Notice that since its movement was 60 km straight east and then from there 40 km straight west, the car is positioned at 20 km to the east of its initial departure point. therefore the vector change in position is a vector 20 km in magnitude, and direction towards the east.

Since it took the car a total of 1.33 hours plus 0.67 hours to reach its final position, the total time elapsed is: 1.33 + 0.67 hours = 2 hours.

Then,the velocity vector has magnitude; 20 km / 2 hours = 10 km/hour

As we mentioned above. the direction of the velocity vector is east.

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Ostrovityanka [42]

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When light strikes a black surface, it is reflected by the surface and

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Transmission in the passing of light through some materials​

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2 years ago
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