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Alik [6]
3 years ago
14

A diver who has a mass of 68 kg climbs to a diving platform that is 7.5 m above the surface of a pool. How much gravitational po

tential energy does the diver have in relation to the pool’s surface?
Physics
2 answers:
noname [10]3 years ago
4 0
The answer is 4998 J.
iren [92.7K]3 years ago
4 0

Answer: 4,998 J

Explanation:

The gravitational potential energy of an object is given by

E_p=mgh

where

m is the mass of the object

g is the gravitational acceleration

h is the heigth of the object with respect to a reference point

In this problem, m=68 kg, g=9.8 m/s^2, and h=7.5 m is the heigth of the diver with respect to the surface of the pool. Therefore, the gravitational potential energy of the diver with respect to the surface of the pool is

E=mgh=(68 kg)(9.8 m/s^2)(7.5 m)=4,998 J

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A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o
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Answer:

a) The proton's speed is 5.75x10⁵ m/s.

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Explanation:

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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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