Answer:
6858.5712 m/s
Explanation:
Given that:
Radius, r
R = 3.20 * 10^3.
Normal force = 0.5 * normal weight
Normal force = Fn ; Normal weight = Fg
Fn = 0.5Fg
Recall:
mv² / R = Fn + Fg
Fn = 0.5Fg
mv² / R = 0.5Fg + Fg
mv² /R = 1.5Fg
mv² = 1.5Fg * R
F = mg
mv² = 1.5* mg * R
v² = 1.5gR
v = sqrt(1.5gR)
V = sqrt(1.5 * 9.8 * 3.2 * 10^3)
V = sqrt(47.04^3)
V = 6858.5712 m/s
Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
Answer:
1.64 * 10^(-5) m
Explanation:
Parameters given:
Angular separation, θ = 0.018 rad
Wavelength, λ = 589 nm = 5.89 * 10^(-7) m
The angular separation when there are 2 slots is given as
θ = λ/2d
where d = separation between slits
d = λ/2θ
d = (589 * 10^(-9))/(2 * 0.018)
d = 1.64 * 10^(-5) m
Answer:
The center of mass of three mass in the x-y plane is located at (1,0.5).
Explanation:
It is given that, a mass of 6 kg is at (0,0), a mass of 4 kg is at (3,0), and a mass of 2 kg is at (0,3). We need to find the center of mass of the system. Center of mass in x direction is :
The center of mass in y direction is :
So, the center of mass of three mass in the x-y plane is located at (1,0.5).
