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3 years ago

If the horse lifts a 50 kilogram mass a vertical distance of 3 meters in 2 seconds, what is the horse’s minimum power output?

Physics

1 answer:

yaroslaw [1]3 years ago

7 0

Answer:

735watts

Explanation:

Power = Force*distance/time

Given

Force = mg = 50 * 9.8'

Force = 490N

distance = 3m

time = 2s

Substitute

Power = 490*3/2

Power = 1470/2

Power = 735Watts

Hence the minimum power output is 735watts

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14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.

sineoko [7]

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

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3 years ago

Explain the diode equation

frosja888 [35]

Answer:

The diode equation gives an expression for the current through a diode as a function of voltage.

Explanation:

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2 years ago

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Which do you think changes the land more frozen water or flowing water?

Maksim231197 [3]

I believe flowing water changes the land because eventually frozen water has to melt to flowing water, right?

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3 years ago

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Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re

lapo4ka [179]

Answer:

$K_{system} = \frac{k}{10}$

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})$

Here, n = 10

Spring constant of each spring = k

Thus,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})$

$\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})$