When light crosses the boundary between layers with different densities, the light is refracted. (A).
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
Answer:
V = 3.17 m/s
Explanation:
Given
Mass of the professor m = 85.0 kg
Angle of the ramp θ = 30.0°
Length travelled L = 2.50 m
Force applied F = 600 N
Initial Speed u = 2.00 m/s
Solution
Work = Change in kinetic energy
Answer:
1201 lbs
Explanation:
Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.
Let the weight of the heart of a mammal be H
And the weight of the total body be B
The linear model that can gives the heart weight in terms of the total body weight will be:
H = 0.005B
B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.
H = 0.005 × 2.402 × 10^5
H = 1201 lbs
Therefore, the weight of the heart of the whale is 1201 lbs
