Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of

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.

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

Response: The spring constant is 25 N/m.

Details:

The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).

The total oscillations are 20 in 4 seconds.

Oscillations per second =

Spring's frequency of vibration is =

The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.

The spring constant is 25 N/m.

A bathroom scale operates under gravitational influence. Typically, a reading is captured when your body applies force onto the scale. Yet in this scenario, as both you and the scale move downwards, your body ceases to press against the scale. Consequently, the result is:

<span>The scale reading will instantly drop to zero</span>

A. The horizontal component of velocity is
vx = dx/dt = π - 4πsin(4πt + π/2)
vx = π - 4πsin(0 + π/2)
vx = π - 4π(1)
vx = -3π

b. vy = 4πcos(4πt + π/2)
vy = 0

c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]

d. m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. To find t, set
vx = π - 4πsin(4πt + π/2) = 0
Then use this to calculate vxmax

h. To determine t, set
vy = 4πcos(4πt + π/2) = 0
Then use this to find vymax

i. s(t) = [x(t)^2 + y(t)^2]^(1/2)

h. s'(t) = d[x(t)^2 + y(t)^2]^(1/2) / dt

k and l. Determine the values for t
d[x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to find both the maximum and minimum speeds.