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LUCKY_DIMON [66]
3 years ago
11

A student was asked to push a large box down the hallway about 35 meters. the task took the student five minutes and a force of

150 newtons. at what rate was the student pushing the box?
Physics
1 answer:
zaharov [31]3 years ago
5 0
Rate of work or Power done by the student can be computed with below formulas.
1. Formula for Work
Work = Force x distance
Force or F = Force applied to the object 
distance or d = displacement

2. Formula for Power
P = Work done / Time it was done or completed

Substituting the values we have,
Work or Work done = 150 N x 35 meters = 5250 J (unit of work).

Lastly computing for the rate we have,
Power or rate of work = 5250 J / 300 seconds (or 5 min) = 17.5 Joule/S or Watts

Answer is 17.5 watts

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A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How mu
puteri [66]

Answer:

84.6 J

Explanation:

The work done by the force is given by

W=Fd cos \theta

where

W is the work done

F = 15 N is the force applied

d = 6.0 m is the displacement

\theta=20^{\circ} is the angle between the force's direction and the displacement

Substituting the numbers into the equation, we find

W=(15 N)(6.0 m)cos 20^{\circ} =84.6 J

3 0
3 years ago
A LETTER FROM THE LORAX
KengaRu [80]

Answer:

Explanation:

You could try to say how helpful they are what they are and what they do

5 0
3 years ago
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A driver in a car traveling at a speed of 21.8m/s sees a cat 101 m away on the road. How long will it take for the car to accele
kobusy [5.1K]
The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:

v^2 = v_0^2 + 2ad\ \to\ a = \frac{-v_0^2}{2d} = -\frac{21.8^2\ m^2/s^2}{2\cdot 99\ m} = -2.4\frac{m}{s^2}

We can clear time in the speed equation:

v = v_0 + at\ \to\ t = \frac{-v_0}{a} = \frac{-21.8\ m/s}{-2.4\ m/s^2} = \bf 9.08\ s

If you find some mistake in my English, please tell me know.
3 0
3 years ago
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A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 324 meters. Assuming no air resist
Elanso [62]

Answer:

8.13secs

Explanation:

From the question weal are given

Height H =324m

Required

time it takes to drop t

Using the equation of motion

H = ut + 1/2gt²

Substitute the given values

324 = 0(t)+1/2(9.8)t²

324 = 1/2(9.8)t²

324 = 4.9t²

t² =324/4.9

t² = 66.12

t = √66.12

t = 8.13secs

Hence the time taken to drop is 8.13secs

4 0
3 years ago
A surveyor is using a magnetic compass 5.6 m below a power line in which there is a steady current of 140 A. (a) What is the mag
ArbitrLikvidat [17]

Answer:

(a) B = 5.6 micro Tesla

Explanation:

Current in the wire, i = 140 A

distance, r = 5 m

The formula for the magnetic field at a distance r due to the current carrying wire

B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B=10^{-7}\times \frac{2\times140}{5}

B = 5.6 x 10^-6 Tesla

B = 5.6 micro Tesla

(b) As the magnetic field of earth at this site is 20 micro tesla so the magnetic field due to current carrying wire interfere the magnetic compass.

4 0
3 years ago
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