Answer:
g₂ = 11 m/s²
Explanation:
The value of free-fall acceleration on the surface of a planet is given by the following formula:
where,
g = free-fall acceleration
G = Universal Gravitational Constant
m = mass of the planet
r = radius of planet
FOR PLANET 1:
--------------------- equation (1)
FOR PLANET 2:
using equation (1):
<u>g₂ = 11 m/s²</u>
Answer:
If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.
Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:
PE = qV
where q =1.6 x 10^-19 C
and V = 1.00 V
Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:
PE* = 0,
KE* = 1/(2mv^2)
Now since
PE + KE = Total energy =PE* + KE*
Therefore,
qV + 0 = 0 + 1/2mv^2
Or
KE = qV = 1.6 10^-19 J
Answer:
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Since we have , v=f×lambda (wavelength). Where v equals 350m/s and wavelength equals 3.80. so it will become f = v/lambda=350/3.80=92.1052Hz
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