We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
Answer:
v= 2413.5 m/s
Explanation:
maximum change of speed of rocket
=(initial exhaust velocity)×ln [(initialmass/finalmass)]
let initial mass= m
final mass = m-m(4/5) = m/5
[since the 80% of mass which is fuel is exhausted]
V-0 = 1500 ln (1/0.2)
V= 1500×1.609 = 2413.5 m/s
therefore, its exhaust speed v= 2413.5 m/s
Answer:
See the explanation below
Explanation:
To better understand this problem, a cylinder sketch is attached before and after the cut, we see that after the cut, the shape of this resembles that of a right triangle.
We can find, the centroid in the xy plane, knowing that the centroid for a triangle is located a third of its base.
In the z axis there is no displacement of the centroid.
Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

hence, B_o = 1.013μT