Answer:
P = 5.22 Kg.m/s
Explanation:
given,
mass of the projectile = 1.8 Kg
speed of the target = 4.8 m/s
angle of deflection = 60°
Speed after collision = 2.9 m/s
magnitude of momentum after collision = ?
initial momentum of the body = m x v
= 1.8 x 4.8 = 8.64 kg.m/s
final momentum after collision
momentum along x-direction
P_x = m v cos θ
P_x = 1.8 x 2.9 x cos 60°
P_x = 2.61 kg.m/s
momentum along y-direction
P_y = m v sin θ
P_y = 1.8 x 2.9 x sin 60°
P_y = 4.52 kg.m/s
net momentum of the body
P = 5.22 Kg.m/s
momentum magnitude after collision is equal to P = 5.22 Kg.m/s
Answer: 2, the nuclear strong force drops to practically nothing at large distances.
Explanation: The protons and neutrons in the nucleus share subatomic particles called pions. This exchange is what keeps the protons and neutrons stuck together in the nucleus. Despite the strong force being the strongest force, it has a very small range. This is because pions have very short lifespans. So, the strong force would have literally no effect at large distances.
Hope that helped! :)
Answer:
U₂ = 400 KJ
Explanation:
Given that
Initial energy of the tank ,U₁= 800 KJ
Heat loses by fluid ,Q= - 500 KJ
Work done on the fluid ,W= - 100 KJ
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take final internal energy =U₂
We know that
Q= U₂ - U₁ + W
-500 = U₂ - 800 - 100
U₂ = -500 +900 KJ
U₂ = 400 KJ
Therefore the final internal energy = 400 KJ
Answer:
B. The escape speed of the Moon is less than that of the Earth; therefore, less energy is required to leave the Moon.
Explanation:
Since the speed required to escape from the gravitational attraction of the Moon is less than the speed required to escape from the gravitational attraction of the Earth, less energy is required to travel from the Moon to the Earth, than is required to travel from the Earth to the Moon. This is because the kinetic energy is directly proportional to the square of the velocity.
The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V
