Answer:
298,220 N
Explanation:
Let the force on car three is T_23-T_34
Since net force= ma
from newton's second law we have
T_23-T_34 = ma
therefore,
T_23-T_34 = 37000×0.62
T_23= 22940+T_34
now, we need to calculate
T_34
Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2
F= ma
So, F= 12×37000×0.62= 22940×12= 275280 N
T_23 =22940+T_34= 22940+ 275280= 298,220 N
therefore, the tension in the coupling between the second and third cars
= 298,220 N
Answer:
0.2932 rad/s
Explanation:
r = Radius = 2 m
= Initial angular momentum = 
= Initial angular velocity = 14 rev/min
= Final angular momentum
= Final angular velocity
Here the angular momentum of the system is conserved

The final angular velocity is 0.2932 rad/s
The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:

where

is the orbital radius of the first satellite

is the orbital radius of the second satellite

is the orbital period of the first satellite

is the orbital period of the second satellite
If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite:
Should be 4x5x4!! If you take 8•15, you get 120!! There are six sides on a rectangular prism!! Thus leaving a value of 20 per side!! And I noted to properly equal out with the maximum volume it should mean 4•5=20!! Thus being 4•4•5
Answer:
The maximum no. of electrons- 
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,

Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:

Maximum number of electrons, n is given by:

where
e = charge on an electron = 
Thus
