Answer:
Therefore the terminal velocity = 1.45 m/s
Explanation:
Terminal velocity: Terminal velocity is the highest velocity of an object when it falls from rest trough a media.
= terminal velocity
w = weight of the object = mg
= drag coefficient=0.80
A= frontal area
= media density = 1.2 kg/m³
m = mass = 8 kg
g= acceleration due to gravity = 9.8 m/s²
Front area = length× breadth
= (18×47)cm²
=846 cm²
Therefore the terminal velocity
=1.45 m/s
Therefore the terminal velocity = 1.45 m/s
The answer is a, series circuit.
Time = distance/speed
Time = 8•10^14m / 3•10^8m/s
Time = about 2.7•10^6 seconds.
That's about 31 Earth days.
Given :
Initial velocity, u = 12.5 m/s.
Height of camera, h = 64.3 m.
Acceleration due to gravity, g = 9.8 m/s².
To Find :
How long does it take the camera to reach the ground.
Solution :
By equation of motion :
Putting all given values, we get :
t = 2.56 and t = −5.116.
Since, time cannot be negative.
t = 2.56 s.
Therefore, time taken is 2.56 s.
Hence, this is the required solution.
