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3 years ago

A weight lifter lifts a 250 Newton weight up 1.5 meters. How much work did they do?

Physics

1 answer:

Rasek [7]3 years ago

6 0

Answer:

375 J

Explanation:

Work done is the product of force and distance moved by the object.

W=F*d

Given that ;

Force= F= 250 N

Distance= d =1.5 m

W= 250 * 1.5 =375 J

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The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3

bagirrra123 [75]

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

v = 117.22 m/s

Approximation to 2 sig. figures gives;

v = 120 m/s

8 0

3 years ago

When a hypothesis is tested many times and supported by data, it becomes a __________. control theory solution conclusion

Kitty [74]

I think it would then become a theory

5 0

3 years ago

Which players are usually the tallest on their team, and stay close to the basket so they can shoot and rebound the ball?

alexdok [17]

Answer:

Center

Explanation:

The center is the tallest player on each team, playing near the basket. On offense, the center tries to score on close shots and rebound. But on defense, the center tries to block opponents' shots and rebound their misses.

7 0

3 years ago

a ball is projected horizontally with a velocity of 5 m per second from the top of a building 19.6 m high how long will the ball

zepelin [54]

Answer:

1.98s

Explanation:

The time taken to hit the ground is given by

h=ut+ 1/2 at^2

but u =0

so we have

h=1/2at^2

making t the subject

t=√2h/g

√2×19.6/10

1.98s

8 0

3 years ago

What is the acceleration of the the object during the first 4 seconds?

AVprozaik [17]

Answer:

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

6 0

3 years ago