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nalin
8 days ago
11

Water is boiled at 1 atm pressure in a 25-cm-internal- diameter stainless steel pan on an electric range. if it is observed that

the water level in the pan drops by 10 cm in 45 min, determine the rate of heat transfer to the pan.
Physics
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A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [3030]

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.

8 0
4 months ago
A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel
kicyunya [3294]

The rocket's acceleration is described here as

a_y = 2.60* t

now recognizing that

\frac{dv}{dt} = 2.60t

we integrate both sides

\int dv = \int 2.60t dt

v = 2.60\frac{t^2}{2}

v = 1.30 t^2

given that the rocket is accelerating for a duration of t = 10 s

thus, we have

v = 1.30 * 10^2

v = 130 m/s

consequently, after t = 10 s, the rocket will achieve a speed of 130 m/s in an upward direction

5 0
2 months ago
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