A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

Response:

2.5kN.m

Details:

Torque relates directly to the pitch diameter

= Ta/Tb= Da/Db

For 120/Tb= 0.25/0.5

This gives Tb= 2.469kN.m, roughly 2.5kN.m

Response:

A)

B)

C)

Clarification:

The question is not fully provided. The complete question was:

A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.  

Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.

Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.

Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we assess the energy balance to determine the work done by the friction:

For part B), we utilize the previously calculated work:

  Rearranging for friction force:

For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:

Rearranging for 'a':

Now, using dynamics:

assuming north-south is along the Y-axis and east-west along the X-axis

X = total X-displacement

from the graph, total displacement in the X-direction is computed as

X = 0 - 20 + 60 Cos45 + 0

X = 42.42 - 20

X = 22.42 m

Y = total Y-displacement

from the graph, total displacement in the Y-direction is computed as

Y = 40 + 0 + 60 Sin45 + 50

Y = 90 + 42.42

Y = 132.42 m

To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding

magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m

Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east