Calculate the deceleration (in m/s2) of a snow boarder going up a 2.65° slope assuming the coefficient of friction for waxed woo
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It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.
From the question given above, the following data were obtained:
Height (H) = 206 m
<h3>Time (t) =? </h3>NOTE: Acceleration due to gravity (g) = 10 m/s²
The time taken for the ball to get to the ground can be obtained as follow:
H = ½gt²
206 = ½ × 10 × t²
206 = 5 × t²
Divide both side by 5
Take the square root of both side
Therefore, it will take 6.42 s for the ball to get to the ground.
Learn more: brainly.com/question/24903556
Answer:
0.075 m
Explanation:
The picture of the problem is missing: find it in attachment.
At first, block A is released at a distance of
h = 0.75 m
above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:
where
is the acceleration due to gravity
is the mass of the block
is the speed of the block A just before touching block B
Solving for the speed,
Then, block A collides with block B. The coefficient of restitution in the collision is given by:
where:
e = 0.7 is the coefficient of restitution in this case
is the final velocity of block B
is the final velocity of block A
is the initial velocity of block B
Solving,
Re-arranging it,
(1)
Also, the total momentum must be conserved, so we can write:
where
And substituting (1) and all the other values,
This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to
where
k = 600 N/m is the spring constant
x is the compression of the spring
And solving for x,
