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Gemiola [76]
3 years ago
10

Which of the following situations would violate the second law of

Physics
1 answer:
Sonja [21]3 years ago
6 0

Answer:

c

Explanation:i don’t know

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Find the kinetic energy ofan electron that moves at half the speed of light to four significant digits.
Vlad1618 [11]

Answer:

Ke=electron kinetic energy=10.24*10^{33} J

Explanation:

The electron has a mass of 9.1*10^{-31} kg

The speed of light in a vacuum is a universal constant with the value 299 792 458 m / s (186 282,397 miles / s), although it is usually close to 3*10^{8} \frac{m}{s}

Kinetic energy (K) is the energy associated with bodies that are in motion, depends on the mass and speed of the body and is calculated using the formula:

K=\frac{1}{2} *m*v^{2}        Equation(1)

K=kinetic energy (J)

m =mass of the body (kg)

v= speed of the body(\frac{m}{s} )

for this problem We replace in the equation (1)

me=9.1*10^{-31} kg = electron mass

ve=1.5*10^{8} \frac{m}{s}=Half the speed of light

=electron speed

We replace in  the  equation (1) :

Ke=\frac{1}{2}*me*ve^{2}

Ke=\frac{1}{2} *9.1*10^{-31} *(1.5*10^{8} )^{2}

Ke=4.55*10^{-31} *(1.5)^{2} *(10^{8} )^{2}

Ke=10.24*10^{33} J

The energy kinetic of the electron is 10.24*10^{33} Joules

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