Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :
(b) Speed of sound in air, v = 343 m/s
It can be given by :
Hence, this is the required solution.
Answer:
v = 7.69 x 10³ m/s = 7690 m/s
T = 5500 s = 91.67 min = 1.53 h
Explanation:
In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:
where,
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Me = Mass of Earth = 5.97 x 10²⁴ kg
r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m
v = orbital speed = ?
Therefore,
<u>v = 7.69 x 10³ m/s</u>
For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.
So, its orbital speed can be given as:
where,
T = Time Period of Satellite = ?
Therefore,
<u>T = 5500 s = 91.67 min = 1.53 h</u>
Answer:
2.84 m/s
Explanation:
At the top position of the circular trajectory, the normal reaction is zero:
N = 0
So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:
where
m is the mass of the water bucket
g = 9.8 m/s^2 is the acceleration of gravity
v is the speed of the bucket
r = 0.824 m is the radius of the circle
Solving for v,