The work function for magnesium is 3.70 ev. what is its cutoff frequency?
1 answer:
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Answer: 1010.92 m/s
Explanation:
According to Newton's law of universal gravitation:
(1)
Where:
is the gravitational force between Earth and Moon
is the Gravitational Constant
is the mass of the Earth
is the mass of the Moon
is the distance between the Earth and Moon
Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to :
(2)
Where is the centripetal acceleration given by:
(3)
Being the orbital velocity of the moon
Making (1)=(2):
(4)
Simplifying:
(5)
Making (5)=(3):
(6)
Finding :
(7)
(8)
Finally:
First, let's put 22 km/h in m/s:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve: