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worty [1.4K]
3 years ago
6

A 63 kg object needs to be lifted 7 meters in a matter of 5 seconds. Approximately how much horsepower is required to achieve th

is task?
A. 0.59 hp
B. 441 hp
C. 1.16 hp
D. 864.36 hp
Physics
1 answer:
Veseljchak [2.6K]3 years ago
4 0
To solve for power you can use this formula:

p =  \frac{mad}{t}

Where: m = mass
             a  = acceleration
             d = distance
             t = time

Since the motion of the object is vertical, the acceleration of the object will be due to gravity. The acceleration due to gravity is 9.8 m /s^2.

So we can now solve for power:

p =    \frac{(63kg)(9.8m/ s^{2} )(7m)}{5s} 


P = 864.36 Watts

The question though is how much horsepower so we need to convert watts into horsepower. 

864.36 watts to hp is 1.159 hp or 1.16 hp the answer is C. 
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(D) The object's kinetic energy is zero.

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Kinetic energy is the energy associated with bodies that are in motion, depends on the mass and speed of the body.

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The kinetic energy, Ek, is measured in joules (J), the mass, m, is measured in kilograms (kg) and the velocity, v, in meters / second (m / s).

Relationship between work and kinetic energy

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc
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Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²    (when the rocket is accelerated upward).

y = y0 +  v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t  (when the rocket has fuel)

v = v0 + g · t   (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

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ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

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