Answer:
The spring stretched by x = 13.7 cm
Explanation:
Given data
Mass = 3 kg
k = 120 
Angle 
= 34°
From the free body diagram
Force acting on the box = mg sin
⇒ F = 3 × 9.81 × 
⇒ F = 16.45 N ------- (1)
Since box is attached with the spring so a spring force also acts on the box.
= k x
= 120 
-------- (2)
The net force acting on the body is given by
Since acceleration of the box is zero so
Put the values from equation (1) & (2) we get
16.45 = 120
x = 0.137 m
x = 13.7 cm
Therefore the spring stretched by x = 13.7 cm
Answer:
Anticlockwise directions
Please mark me Brainliest to help me
I’m pretty sure the answer is C. Any change of state or movement requires energy
Answer:
The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.
Explanation:
- The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
- The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
- During terminal velocity, we can represent mathematical equation as;
Buoyancy force + drag force = Gravity
F = 52000 N
m = 1060 kg
a= F/m = 52000 N/1060 kg = 49.0566 m/s^2
