31.3m/s
Explanation:
Given parameters:
Mass of rock = 40kg
Height of cliff = 50m
Unknown:
Speed of rock when it hits ground = ?
Solution:
We are going to use the appropriate motion equation to solve this problem
The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.
Using;
V² = U² + 2gH
V = unknown velocity
U = initial velocity = O
g = acceleration due to gravity = 9.8m/s²
H = height of fall
since the initial velocity of the bodyg is 0
V² = 2gH
V= √2gH = √2 x 9.8 x 50 = 31.3m/s
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Velocity brainly.com/question/4460262
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Newton's first law of motion states that an object at rest tends to stay at rest, while an object in motion tends to stay in motion unless an external force acts upon it. This law appears in basketball when the player is shooting the ball. When the player is holding the ball, the ball is at rest but when a player shoots the ball, they use force to throw the ball in the hoop.
Answer:
Distance of 400m.
Explanation:
Use your kinematics equation to solve for distance (we can use kinematics b/c acceleration is constant).
d = (initial velocity x time) + 1/2 at^2
d = (20 x 10) + 1/2 (4) (10)^2
d = 200 + 200
d = 400 m
Answer:
When a physical change in a sample occurs, composition of the sample does not change. It stays the same. It stays the same. Also, the properties of the sample will still be the same
