A baseball travels 150 km/h. What does this measurement represent?
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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.
The diameter of the disks is 1.32 cm.
If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,
Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.
The potential difference V between the disks separated by a distance d is given by,
here, E is the electric field.
Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.
The capacitance C of the capacitor is given by,
The capacitance of a parallel plate capacitor is given by,
Here, ε₀ is the permittivity of free space and A is the area of the disks.
Rewrite the expression for A.
the area A of the disks is given by,
Here, D is the diameter of the disk.
The diameter of each disc is found to be 1.32 cm.