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3 years ago

A particle moving on a circle has a velocity of 5 m/s and a normal acceleration of 10 m/s^2. What is the radius of the circle?

Physics

1 answer:

dybincka [34]3 years ago

3 0

Answer:

Radius of the circle will be 2.5 m

Explanation:

We have given velocity of particle moving in the circle v = 5 m/sec

Acceleration of particle in the circle $a=10m/sec^2$

We have to find the radius of the circle

We know that acceleration is given by $a=\frac{v^2}{r}$

So $10=\frac{5^2}{r}$

$r=\frac{25}{10}=2.5m$

So radius of the circle will be 2.5 m

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A)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.

arsen [322]

The solution for the acceleration of gravity is given as

$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

This is further explained below.

<h3>What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?</h3>

Generally,

Mass of earth $M=5.97 \times 10^{24} \mathrm{~kg}$

Radius of earth $R=6371 \mathrm{~km}$

Gravitational const. $G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$

height $h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$

$R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}$

In conclusion, acceleration due to gravity at this point will be

$g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

for $h_{2}=7900 \mathrm{~km}$

$R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

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