Explanation:
We'll call the radius r and the diameter d:
We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.
The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).
It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.
Hopefully I understood the question. If yes, that's the answer.
6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
7. Same idea as question 2.
First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N.
10. First find the force of block B to the right due to its acceleration:
F = ma
F = (24)(0.5)
F = 12N
So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:
The sum of the forces in the leftward and rightward direction for block B must equal 12N.
75 - x = 12
x = 63N
So the force of friction of block A on block B is 63N to the left.
Answer: 
Explanation:
Given
mass of ball m=10 kg
It is placed at a height of 150 m
It is dropped from the height and allowed to free fall for 40 m
Velocity acquired by the ball during this fall is given by 
Insert u=0, a=g
Kinetic energy at this instant
Answer:
Given that
D= 4 mm
K = 160 W/m-K
h=h = 220 W/m²-K
ηf = 0.65
We know that
For circular fin
m = 37.08
By solving above equation we get
L= 36.18 mm
The effectiveness for circular fin given as
ε = 23.52
Answer:
The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.
Explanation:
As from the given data
the length of the rope is given as l=30 m
the stretched length is given as l'=41m
the stretched length required is give as y=l'-l=41-30=11m
the mass is m=95 kg
the force is F=380 N
the gravitational acceleration is g=9.8 m/s2
The equation of k is given by equating the energy at the equilibrium point which is given as
Here
m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so
Now the force is
or
So here F=380 N, k=630.92 N/m
So the distance is 0.602 m
