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Helga [31]
3 years ago
6

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis

tance of 1.1 m in a time of 1.3 s. The readout on the display indicates that the average power you are producing is 92 W. What is the magnitude of the force that you exert on the handle?
Physics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

The magnitude of force exerted on the handle is 108.73 N

Explanation:

To determine the magnitude of force exerted, we will use the formula relating Power and Force.

Power is the rate at which work is done. Power can be calculated from the formula

Power = Work / Time

But, Work = Force × Distance

Hence,

Power is given by the formula

P = \frac{F \times s}{t}

Where P is the Power

F is the force

s is the distance

and t is the time

From P = \frac{F \times s}{t},

Then we can write that

F = \frac{P \times t}{s}

From the question,

Distance, s = 1.1 m

Time, t = 1.3 s

Power, P = 92 W

Putting these values into the formula, we get

F = \frac{92 \times 1.3}{1.1}

F = \frac{119.6}{1.1}

F = 108.73N

Hence, the magnitude of force exerted on the handle is 108.73 N.

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How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is
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1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

<h3>What is Speed ?</h3>

Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s

In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.

What are the parameters to consider ?

The parameters are;

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Speed = distance S ÷ Time t

Convert kilometer to meter by multiplying it by 1000

C = S/t

3 × 10^{8} =  3.85 × 10^{8} / t

Make t the subject of formula

t = 3.85 × 10^{8} / 3 × 10^{8}

t = 1.2833

t = 1.3 s

Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

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2 years ago
For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener
Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

v=u+at

where:

v is the velocity at time t

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

u=0 (the object starts from rest)

a=9.8 m/s^2

Therefore, the velocity after

t = 2 s

is:

v=0+(9.8)(2)=19.6 m/s (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

KE=\frac{1}{2}(3)(19.6)^2=576 J

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

a=g=9.8 m/s^2 (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

s=0+\frac{1}{2}(9.8)(2)^2=19.6 m

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

KE'=\frac{1}{2}Mv^2

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

KE'=\frac{1}{2}(6)(19.6)^2=1152 J

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

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If the mass of an object is 176 and Net force is 50, what would be the Acceleration?
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50/176 = a ≈ 0.28 m/s²

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