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
The answer is point b because vertical velocity is zero at the maximum height
The answer is 60 km. I hope it helps i dont know if this is right or wrong.
Answer:
The friction force acting on the object is 7.84 N
Explanation:
Given;
mass of object, m = 4 kg
coefficient of kinetic friction, μk = 0.2
The friction force acting on the object is calculated as;
F = μkN
F = μkmg
where;
F is the frictional force
m is the mass of the object
g is the acceleration due to gravity
F = 0.2 x 4 x 9.8
F = 7.84 N
Therefore, the friction force acting on the object is 7.84 N
<h3>
Answer:</h3>
117.6 Joules
<h3>
Explanation:</h3>
<u>We are given;</u>
- Force of the dog is 24 N
- Distance upward is 4.9 m
We are required to calculate the work done
- Work done is the product of force and distance
- That is; Work done = Force × distance
- It is measured in Joules.
In this case;
Force applied is equivalent to the weight of the dog.
Work done = 24 N × 4.9 m
= 117.6 Joules
Hence, the work done in lifting the dog is 117.6 Joules
