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defon
3 years ago
12

An object is dropped from a height h. during the final second of its fall, it traverses a distance of 41.2 m. what was h?

Physics
1 answer:
Leya [2.2K]3 years ago
3 0
41.2 = h-1/2g(t-1)^2 
<span> {-h = -1/2gt^2-1/2g+g*t-41.2 
</span><span> {h = 1/2gt^2 
</span><span> summing them up 
</span><span> 0 = -1/2g+g*t-41.2 
</span><span> 41.2 +4.9 = g*t 
</span><span> t = 46.1/9.8 = 4.70 sec 
</span><span> h = 1/2gt^2 =4.9*(4.70^2) = 108.241 m </span>
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Answer:

3.1\cdot10^{23}\:\mathrm{kg}

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

g_P=G\frac{m}{r^2}., where g_P is acceleration due to gravity at the planet's surface, G is gravitational constant 6.67\cdot 10^{-11}, m is the mass of the planet, and r is the radius of the planet.

Since acceleration due to gravity is given as m/s^2, our radius should be meters. Therefore, convert 2400 kilometers to meters:

2400\:\mathrm{km}=2,400,000\:\mathrm{m}.

Now plugging in our values, we get:

3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2},

Solving for m:

m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}.

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Answer:

a= - 0.79 m/s²

Explanation:

Given that

Speed ,u = 20 mi/h

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1 mi/h= 0.44 m/s

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Lets take the acceleration of the car = a m/s²

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Therefore the acceleration will be - 0.79 m/s².

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