We have:
Total Energy: KE + GPE
KE (Kinetic Energy) =
GPE (Gravitational Potential Energy) =
Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²
Formula:
Total Energy: KE + GPE
Solving:
Answer:
F = (913.14 , 274.87 )
|F| = 953.61 direction 16.71°
Explanation:
To calculate the resultant force you take into account both x and y component of the implied forces:
Thus, the net force over the body is:
Next, you calculate the magnitude of the force:
and the direction is:
Answer:
B.convection currents in the mantle
Explanation:
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
