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
long does it take to boil away 2.40 kg of the liquid.
Boiling point of He is 
Latent heat of vapourization 
Power of electrical heater 
mass of liquid is 
amount of heat required to boil
Power 
The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.
The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.
Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.
For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744
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A = (v - u) / t
a = acceleration
v = final velocity
u = initial velocity
t = time
45 = (v - 300) / 10
45 × 10 = v - 300
450 + 300 = v
v = 750 m/s
Hope this helps!
P.S. Let me know if you need an explanation
Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
