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2 years ago

A weightlifter lifted 800 Newtons to a height of 2.5 meters. What is the power output if this was

Physics

1 answer:

stiv31 [10]2 years ago

5 0

The power required is 1.6 kW.

Answer:

Explanation:

Power is defined as the amount of work done on any object for a given time interval. In other words, power is the amount of force required to move an object in a given period of time.

Power = Work done / time taken for that work done.

Here the force is given as 800 N and the displacement is given as 2.5 m, while the time required for the displacement is given as 1.22 seconds.

So the power will be ratio of the product of force acting on the weightlifter, displacement of the weight to the time taken for that displacement.

Power = (800×2.5)/1.22 =1639 W

So the power required is 1.6 kW.

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baherus [9]

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$6.673 \times {10}^{ - 11}$

And unit is Nm^2/kg^2

8 0

2 years ago

Read 2 more answers

How much work do you do on a 15 N book in lifting it straight up for a distance of 0.40 meters?

astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of 0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

$W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J$

So, the magnitude of work done in lifting the book is 6 joules.

7 0

3 years ago

A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6

storchak [24]

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = - 4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = $\sqrt{(4.6^2+7.1^2)}$

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

= 49.5 m

6 0

2 years ago

Read 2 more answers

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$

3 0

3 years ago

Read 2 more answers

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is los

Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100

and since mass and gravity are constant so it leaves us with just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction

b) Here use the equation vf^2=vi^2+2gd

where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable
we know that at maximum height or peak, the velocity is 0 so vf is 0

therefore,

vi =sqrt(-2gd) 
vi =sqrt(-2x-9.81x1.5) 
vi =5.4 m/s

c) The energy was converted to heat due to friction with the air and the ground.

6 0

3 years ago