Answer:
42.05 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Height (h) = 239 m
Acceleration due to gravity (g) = 3.7 m/s²
Final velocity (v) =?
The velocity with which the camera hits the ground can be obtained as follow:
v² = u² + 2gh
v² = 0² + 2 × 3.7 × 239
v² = 0 + 1768.6
v² = 1768.6
Take the square root of both side
v = √(1768.6)
v = 42.05 m/s
Therefore, the velocity with which the camera hits the ground is 42.05 m/s
Answer:
It has no effect on the amplitude.
Explanation:
When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.
Answer:
b) 2.2 kg
Explanation:
Net force acting on an object is the sum of the two forces acting on the body.
The net force is calculated using the parallelogram law of vectors.
F =
Here A = 20 N , B = 35 N and θ =80°
Net Force = F = 43.22 N
Acceleration = a = 20 m/s/s
Since F = ma, m = F/a = 43.22 / 20 = 2.161 kg = 2.2 kg
Answer:
C = (7 + 8) i + (5 - 1) j adding vectors
C = 15 i + 4 j
theta = arctan (4 / 15) = 14.9 deg
Note that this is the same as adding x and y components
Answer:
Static friction
Explanation:
Static friction is the friction force that acts on objects that are not moving.