A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g 5 9.80 m/s2. Find (a) the period of the pend
1 answer:
Answer:
(a) 1.5 second
(b) 0.56 m
Explanation:
Pendulum makes 120 oscillations in 3 min that means in 180 seconds
time taken by the pendulum to complete one oscillation is called time period.
(a) So, the time period is 180 / 120 = 1.5 second
T = 1.5 second
Thus, the time period of the pendulum is 1.5 second.
(b) g = 9.8 m/s^2
The formula for the time period is given by
Where, L be the length of pendulum
L = 0.56 m
Thus, the length of the pendulum is 0.56 m .
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The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:
where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
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Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :
F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.