Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen
Neglect air resistance.
The PE (potential energy) is
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J
The PE is converted into KE (kinetic energy) after the fall.
Therefore the PE decreased by 196 J ≈ 200 J
Answer: d. It has decreased by 200 J
Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:

Finally, you replace the values of all parameters in the previous equation for I:

The moment of inertia of the door around the hinges is 2 kgm^2
Answer:
(A) L = 115.3kgm²/s
(B) dL/dt = 94.1kgm²/s²
Explanation:
The magnitude of the angular momentum of the rock is given by the foemula
L = mvrSinθ
We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.
Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =
115.3 kgm²/s
(B) The magnitude of the rate of angular change in momentum is given by
dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²
Answer:
The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J
Explanation:
When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.
The relationship between the maximum kinetic energy (
) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:
= h(f − f₀)
= hf - hf₀
E is the energy of the absorbed photons: E = hf
ϕ is the work function of the surface: ϕ = hf₀
= E - ϕ
Frequency f = 8.12×10¹⁸ Hz
Maximum kinetic energy
= 4.16×10⁻¹⁷ J
Speed of light c = 3 x 10⁸ m/s
Planck's constant h = 6.63 × 10⁻³⁴ Js
E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸
E = 53.8356 x 10⁻¹⁶ J
from
= E - ϕ ;
ϕ = E - 
ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷
ϕ = 53.4196 x 10⁻¹⁶ J
The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J