9.The force of gravity between two asteroids is 10,000 newtons (N).
1 answer:
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Answer:
1.92 x 10⁻¹²J
Explanation:
The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e
F = ma
Where;
a = [v = linear velocity, r = radius of circular path]
=> F = m ------------(i)
We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;
F = q v B sin θ ----------(ii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = the angle between the velocity and the magnetic field.
Combining equations (i) and (ii) gives
m = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]
=> m = q v B sin 90°
=> m = q v B
Divide both side by v;
=> m = qB
Make v subject of the formula
v =
From the question;
B = 1.25T
m = mass of proton = 1.67 x 10⁻²⁷kg
r = 0.40m
q = charge of a proton = 1.6 x 10⁻¹⁹C
Substitute these values into equation(iii) as follows;
v =
v = 4.79 x 10⁷m/s
Now, the kinetic energy, K, is given by;
K = mv²
m = mass of proton
v = velocity of the proton as calculated above
K =
K = 1.92 x 10⁻¹²J
The kinetic energy is 1.92 x 10⁻¹²J
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s