The centripetal force on the car as it goes around the second curve is twice that compared to the first.
What is Centripetal force?
It is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation.
The formula of Centripetal force is:
F(c) = (m* v^2) / r
Here,
At the first curve,
The curve of radius = r
The constant speed = v
At the second curve,
The car speed (v')= 2 v
The radius of the curve (r')=2 r
According to the formula of centripetal Force:
As the car goes around the second curve,
F'(c) = m*v'^2 / r'
F'(c) = m* (2*v)^2 / 2r
F'(c) = 2* F
Thus,
The centripetal force on the car as it goes around the second curve is twice that compared to the first.
Learn more about centripetal force here:
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Answer: hello question b is incomplete attached below is the missing question
a) attached below
b) V = 0.336 ft/s
Explanation:
Elongation ( Xo) = 16/ 7 feet
mass attached to 4-foot spring = 16 pounds
medium has 9/2 times instanteous velocity
<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>
The motion is an underdamped motion because the value of β < Wo
Wo = 3.741 s^-1
attached below is a detailed solution of the question
1) The total mechanical energy of the rock is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
is the mass,
is the gravitational acceleration and
is the height.
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
Answer:
We cannot place three forces of 5g, 6g, and 12g in equilibrium.
Explanation:
Equilibrium means their sum must be zero.
Here the forces are 5g, 6g, and 12g.
For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.
Here
Magnitude of largest force = 12 g
Sum of magnitudes of other forces = 5g + 6g = 11g
Magnitude of largest force > Sum of magnitudes of other forces
So this forces cannot form equilibrium.
We cannot place three forces of 5g, 6g, and 12g in equilibrium.