<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.
The value of the critical angle is given by:

when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
With time, momentum increases as it builds speed assuming their is nothing in the way to stop it. Based on the graph, you can see that example being displayed as the line on the graph gets higher
Answer:
e. The net magnetic flux in this case would be equal to zero.
Explanation:
As per Gauss law of magnetism we need to find the net magnetic flux through a closed loop
here we know that net magnetic flux is the scalar product of magnetic field vector and area vector
so here we have
= net magnetic flux
since we know that magnetic field always forms closed loop so if we find the integral over a closed loop
then in that case the value of the close integral must be zero
so correct answer would be
e. The net magnetic flux in this case would be equal to zero.
Proceed with caution before crossing