The formula is Ke = 1/2 m v^2
The two of them together have a Ke of mv^2. So you either increase m or v. That's what makes the problem difficult. He can do D or B. We have to choose.
A is no solution. The Ke goes down because Paul loses Ivan's mass.
C is out of the question 3 meters/sec is a big reduction from 5 m/s. So now what do we do about B and D?
The question is what does the third person add. The tandoms I've peddled only allow for 1 or 2 people to add to the motion. So the third person only adds mass. He does not have a v that he is contributing to. To say that he is going 5m/s is true, but he's not contributing anything to that motion.
I pick B, but it is one of those questions that the correctness of it is in the head of the proposer. Be prepared to get it wrong. Argue the point politely if you agree with me, but back off as soon as you have presented your case.
B <<<<====== answer.
Answer:
F = 2.01*10^-16N -^k
Explanation:
In order to calculate the magnetic force perceived by the bee, you use the following formula:
(1)
q: charge of the bee = 1pC = 1*10^-12 C
The average speed of a bee and the magnetic field of the earth are:
v = 6.70m/s
B = 30*10^-6 T
The bee is flying to the west (-^i). You consider that the magnetic field direction is to the north (^j). Then, the direction of the magnetic force is:
-^i X ^j = -^k
You replace the values of the parameters in the equation (1), in order to calculate the magnitude of the force:

The magnetic force perceived by the bee is 2.01*10^-16N in the -^k direction, that is, toward the ground
We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2
Answer:
1) 64.2 mi/h
2) 3.31 seconds
3) 47.5 m
4) 5.26 seconds
Explanation:
t = Time taken = 2.5 s
u = Initial velocity = 0 m/s
v = Final velocity = 21.7 m/s
s = Displacement
a = Acceleration
1) Top speed = 28.7 m/s
1 mile = 1609.344 m

1 hour = 60×60 seconds


Top speed of the cheetah is 64.2 mi/h
Equation of motion

Acceleration of the cheetah is 8.68 m/s²
2)

It takes a cheetah 3.31 seconds to reach its top speed.
3)

It travels 47.5 m in that time
4) When s = 120 m

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds