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3 years ago

2. A 75 kg runner accelerates from 0.00 m/s to 10.0 m/s in 1.5 seconds.

1 answer:

3 0

A :-) A.a.) Given - u = 0.00 m/s
v = 10.0 m/s
t = 1.5 sec
m = 75 kg
Solution -
a = v - u by t
a = 10 - 0 by 1.5
a = 10 by 1.5
a = 6.6 m/s^2

A.b.) sorry ! I don’t no how to do this question

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In a closed energy system, energy is?

Alisiya [41]

Answer:

Explanation:

A closed system can exchange energy but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter.

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2 years ago

The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.

kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

$P = 300\ N \ \times \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\
P = 300 \ N \times 22.22 \ m/s\\\\
P = 6,666 \ W$

The power of the engine at the given efficiency is calculated as follows;

$E = \frac{P_{out}}{P _{in}} \times 100\%\\\\
60\% = \frac{P_{out}}{6,666} \times 100\%\\\\
0.6 = \frac{P_{out}}{6,666} \\\\
P_{out} = 3,999.6 \ W$

Learn more about efficiency here: brainly.com/question/15418098

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2 years ago

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harina [27]

Answer:

Energy

Explanation:

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3 years ago

The GPE of a 70kg person standing on a chair 1m off the ground is how many joules?

Artemon [7]

GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.

U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J

U = 686.7J

Hope this helps!

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3 years ago

4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c

Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

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3 years ago