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Kitty [74]
2 years ago
15

If an astronaut can jump straight up to a height of 0.6 m on earth, how high could he jump on the moon?

Physics
1 answer:
Zinaida [17]2 years ago
4 0

Answer:

Explanation:

Given

Person on earth can jump to a height s=0.6\ m

initial velocity is u

using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

final velocity is zero

s=\frac{u^2}{2g}

0.6=\frac{u^2}{2g}-----1

On moon surface acceleration due to gravity is \frac{1}{6}[/tex] th of earth gravity

so height attained is given by

h=\frac{u^2\times 6}{2\cdot g}-----2

divide 1 and 2 we get

h=6\times 0.6=3.6\ m                    

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q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

K = \frac{1}{2}mv^{2}

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

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