A bowling ball moving with a velocity of 5V to the right collides elastically with a beach ball moving at a velocity 2V to the l
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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
An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.
What is the tension between her ears?
Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?
Answer:
The tension between the ears = 2.07 KN
The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.
Explanation:
Given that:
Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m
Mass of the black hole (m) =
where :
Then; we have:
Schwarzchild radius of the black hole
r - 15.0 km
Mass of each ear
Farther distance between one ear and the black hole (d) = 6.0 cm
= 0.06 m
Closer distance between the other ear and the black home is (d) 6.0 cm
= 0.6 cm
NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .
The net force on the ear closer to the black hole will be:
The net force on the ear farther to the black hole is :
Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:
The tension between the ears = 2.07 KN
The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.