Answered

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What is the efficiency of a machine that lifts a load of 12.0 kg a vertical distance of 5.00
m in 30.0 s after being supplied with 70.0 W of power? Explain the steps you take to get the answer.

Sagot :

rafaleo84

Answer:

Efficency = 28%

Explanation:

We have the load has a weight of:

[tex]w =m*g\\w =12*9.81\\w = 117.72 [N][/tex]

Work in physics is defined as the product of force (weight) by distance.

[tex]W=w*d[/tex]

where:

W = work [J]

w = weight = 117.72[N]

d = distance = 5 [m]

[tex]W=117.72*5\\W=588.6[J][/tex]

Now power is defined as the relationship of work at a certain time.

[tex]P=W/t\\[/tex]

where:

P = power [W]

W = work = 588.6[J]

t = time = 30[s]

[tex]P=588.6/30\\P=19.62[W][/tex]

Now the efficiency of a machine is defined as the power output over the power input to the machine. The power input should always be greater than the power output.

[tex]efficency = \frac{Power_{out}}{Power_{in}}\\Efficeny = \frac{19.62}{70} \\Efficency = 0.28\\Efficency = 28%[/tex]

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