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Margarita [4]
3 years ago
13

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
Physics
2 answers:
Natali [406]3 years ago
3 0
Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:

W = Fd

where:

F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration

Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.

W = Fd = mad = 63(9.81)(7) = 4326.21 Joules

Now that we have work, we can calculate power.

P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts

Convert watts to horsepower (1 hp = 745.7 watts)

P = 865.242 watts (1hp/745.7 watts) = 1.16 hp

Vesna [10]3 years ago
3 0

work = mgh

power = mgh/t = 63•9.8•7/5 = 864 watts

1 HP = 746 watts

so that is 834/746 = 1.2 HP

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3 years ago
With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

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Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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3 years ago
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