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Mademuasel [1]
3 years ago
14

What is the average power output pf a weight lifter who can lift 250 kg 2.0 m in 2.0 s

Physics
1 answer:
liq [111]3 years ago
8 0

Answer:

Power = 2.45Kw or 2450 Watts.

Explanation:

<u>Given the following data;</u>

Mass, m = 250kg

Height, h = 2m

Time, t = 2secs

We know that acceleration due to gravity, g is equal to 9.8m/s²

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

But Energy = mgh

Substituting into the equation, we have

Power = \frac {mgh}{time}

Power = \frac {250*9.8*2}{2}

Power = \frac {4900}{2}

Power = 2450 Watts

To convert to kilowatt (Kw), we would divide by 1000

Power = 2450/1000

Power = 2.45Kw.

Therefore, the average power output of the weightlifter is 2.45 Kilowatts.

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What is the acceleration experienced by a car that takes 10s to reach 27m/s from rest
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These are nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the oz
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Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,
makkiz [27]

Answer: \theta=cos^{-1}0.991=7.69^o

The following vectors have been given: \vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}

The angle between these two vectors can be found by:

cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\&#10;||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}

\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54

||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}

cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o

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3 years ago
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