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Allisa [31]
2 years ago
9

A car engine supplies 2.0 x 103 joules of energy during the 10. seconds it takes to accelerate the car along a horizontal surfac

e. What is the average power
developed by the car engine while it is accelerating?
1. 2.0 x 10^1 w
2. 2.0 x 10^2 w
3. 2.0 x 10^3 W
4. 2.0 x 10^4 w
Physics
1 answer:
andrew11 [14]2 years ago
7 0

Answer: 2. 2.0*10^2 W

Explanation:

Power = Work/Time

Power = (2.0*10^3) Joules/10 seconds

Power = 2.0*10^2 Watts

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Mama L [17]

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about – 0.7

Explanation:

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3 years ago
What is the equivalent resistance of the
BigorU [14]

Answer:

Approximately 111\; {\rm \Omega}.

Explanation:

It is given that R_{1} = 200\; {\Omega} and R_{2} = 250\; {\Omega} are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage V.

Since R_{1} and R_{2} are connected in parallel, the voltage across the two resistors would both be V. Thus, the current going through the two resistors would be (V / R_{1}) and (V / R_{2}), respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: I = (V / R_{1}) + (V / R_{2}).

In other words, if the voltage across this circuit is V, the total current in this circuit would be I = (V / R_{1}) + (V / R_{2}). The (equivalent) resistance R of this circuit would be:

\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}.

Given that R_{1} = 200\; {\Omega} and R_{2} = 250\; {\Omega}:

\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}.

7 0
2 years ago
Cesar and Jill went to a field to play soccer. As the ball downward toward Jill, Jill used her foot to kick the ball and keep it
Kobotan [32]

Answer:

D

Explanation:

8 0
3 years ago
How would you find the average speed of a cyclist throughout an entire race
pishuonlain [190]
<span>Velocity, you divide distance/time </span>
4 0
3 years ago
An object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34
Travka [436]

The initial position of the object was found to be 134.09 m.

<u>Explanation:</u>

As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

Initial position of the object = Final position-Displacement

Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.

Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

4 0
3 years ago
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