The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:

(1)
The work done is equal to the elastic energy stored by the compressed spring:

where

is the spring constant and

is the compression of the spring. If we substitute the numbers, we find:

And now we can use eq.(1) to calculate the average power output:
The kayaker has velocity vector
<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )
<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )
and the current has velocity vector
<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )
<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )
The kayaker's total velocity is the sum of these:
<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>
That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that
tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s) → <em>θ</em> ≈ 18.4º
or about 18.4º north of east.
Kinetic energy is a form of energy that an object or particle has by reaction of its motion.
Answer: A
Explanation:
From the question, the given parameters are given.
Mass M = 30kg
Radius r = 2 m
Coefficient of static friction μ = 0.8
Coefficient of kinetic friction μ = 0.6
Kinetic friction Fk = μ × mg
Fk = 0.6 × 30 × 9.8
Fk = 176.4 N
The force acting on the merry go round is a centripetal force F.
F = MV^2/r
This force must be greater than or equal to the kinetic friction Fk. That is,
F = Fk
F = 176.4
Substitute F , M and r into the centripetal force formula above
176.4 = (30×V^2)/2
Cross multiply
352.8 = 30V^2
V^2 = 352.8/30
V = sqrt (11.76) m/s
V = 5.24 m/s
Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately
Answer:
There is pressure on both the in and outsides of the balloon: high inside and low outside, respectively. In terms of air mass, high pressure has greater activity and tends to move upward (balloon rises); whereas low presssure, being slower, descends (balloon descends).
Explanation: