Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a
rea density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?
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Response:
0.9 cm
Clarification:
The following illustrates the calculation of the combined rod's length increase:
As established
Length increase = expansion of aluminum rod + expansion of steel rod
= 0.9 cm
We simply summed the expansions of both the aluminum and steel rods to determine the overall increase in the joined rod's length, which must be factored in
Answer
Given data:
height of the dam = 15 m
effective area for water flow = 2.3 x 10⁻³ m²
Applying the principle of energy conservation:
v = 17.15 m/s
water discharge
Q = A V
Q = 2.3 x 10⁻³ x 17.15
Q = 0.039 m³/s
Response:
E = ρ ( R1²) / 2 ∈o R
Clarification:
Provided information
Two cylinders are aligned parallel
Distance = d
Radial distance = R
d < (R2−R1)
To determine
Express the response using the variables ρE, R1, R2, R3, d, R, and constants
Solution
We have two parallel cylinders
therefore, area equals 2 R × l
And we apply Gauss's Law
EA = Q(enclosed) / ∈o......1
Initially, we calculate Q(enclosed) = ρ Volume
Q(enclosed) = ρ ( R1² × l )
Thus, inserting all values into equation 1
produces
EA = Q(enclosed) / ∈o
E(2 R × l) = ρ (
R1² × l ) / ∈o
This simplifies to
E = ρ ( R1²) / 2 ∈o R