The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e
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Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency,
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f =
f =
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
Answer:
h = 1.22 m
Explanation:
Given:
Pressure in the vein = 12200 Pa
Specific gravity of the liquid = 1.02
now,
the pressure due to a fluid is given as:
P = pgh
where,
P is the pressure,
ρ is the density of fluid = specific gravity x density of water = 1.02 x 1000 kg/m³
ρ = 1020 kg/m³
g is the acceleration due to the gravity = 9.81m/s²
h is the height
thus,
h = P/pg =