All categories

1 year ago

If a runner exerts 457 j of work to make 321 w of power then how long did it take the runner to do the work

Physics

1 answer:

MA_775_DIABLO [31]1 year ago

5 0

The time taken by the runner to do the work is 1.42 seconds.

Given the data in the question;

Work done; $W = 457J$

Power; $P = 321W$

Time elapsed; $t = \ ?$

<h3>Power</h3>

Power can be simply referred to as the quantity of energy transferred per unit time.

It is expressed as;

$P = \frac{W}{t}$

Where W is work done and t is time elapsed.

To determine the time it took the runner to do the work, we substitute our given values into the expression above.

$P = \frac{W}{t} \\\\t = \frac{W}{P} \\\\t = \frac{457J}{321W} \\\\t = \frac{457 kgm^2/s^2}{321 kgm^2/s^3}\\\\t = 1.42s$

Therefore, the time taken by the runner to do the work is 1.42 seconds.

Learn more about Power and Work: brainly.com/question/2962104

You might be interested in

A cyclist accelerates from 0m/s to 8m/s in 3 seconds.

garik1379 [7]

Answer:

the rate of acceleration is 2.6666(and so on) if thats your question

Explanation:

5 0

2 years ago

What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?

zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

6 0

3 years ago

Read 2 more answers

A 5,257 kg rocket blasts off to the moon with an acceleration of 76 m/s ^2 what is the net force on the rocket

frutty [35]

Newton's subsequent law expresses that power is corresponding to what exactly is needed for an object of consistent mass to change its speed. This is equivalent to that item's mass increased by its speed increase.

We use Newtons, kilograms, and meters each second squared as our default units, albeit any proper units for mass (grams, ounces, and so forth) or speed (miles each hour out of every second, millimeters per second², and so on) could unquestionably be utilized also - the estimation is the equivalent notwithstanding.

Hence, the appropriate answer will be 399,532.

Net Force = 399532

7 0

3 years ago

Ill give 15 points to the one who helps me with this

alexandr402 [8]

Machenical number one

6 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago