Answer:
1.7 m/s²
Explanation:
d = length of the ramp = 13.5 m
v₀ = initial speed of the skateboarder = 0 m/s
v = final speed of the skateboarder = 7.37 m/s
a = acceleration
Using the equation
v² = v₀² + 2 a d
7.37² = 0² + 2 a (13.5)
a = 2.01 m/s²
θ = angle of the incline relative to ground = 29.9
a' = Component of acceleration parallel to the ground
Component of acceleration parallel to the ground is given as
a' = a Cosθ
a' = 2.01 Cos29.9
a' = 1.7 m/s²
The vertical velocity of the projectile upon returning to its original is 17. 74 m/s
<h3>
How to determine the vertical velocity</h3>
Using the formula:
Vertical velocity component , Vy = V * sin(α)
Where
V = initial velocity = 36. 6 m/s
α = angle of projectile = 29°
Substitute into the formula
Vy = 36. 6 * sin ( 29°)
Vy = 36. 6 * 0. 4848
Vy = 17. 74 m/s
Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s
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1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J
The elastic potential energy of a spring is given by

where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:

In our problem, initially the spring is uncompressed, so

. Therefore, the work done by the spring when it is compressed until

is

And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
Answer:
The centripetal force on body 2 is 8 times of the centripetal force in body 1.
Explanation:
Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.