Answer: 6.125 ft
Explanation:
If this dish has the form of a concave upward parabola and its vertex 
is at the origin, its corresponding equation is:
Where:
is the radius, which can be found by dividing the diameter 
by half. Hence 
is the depth
is the vertex of the parabola, where its base is
Finding :
Finally:
This is where the the receiver should be placed
Answer:
The temperature of the Aluminium plate 44.84⁰C
Explanation:
Number of transistors = 4
Since the heat dissipated by each transistor is 12W
Total heat dissipated, Q = 4 * 12 = 48 W
Q = 48 W
Cross sectional Area of the Aluminium plate, A = 2(l * b)
l = Length of the aluminium plate = 22 cm = 0.22 m
b = width of the aluminium plate = 22 cm = 0.22 m
A =2( 0.22 * 0.22 )
A = 0.0968 m²
From the heat balance equation, Q = hAΔT
h = 25 W/m²·K
A = 0.0968 m²
ΔT = T - T(air)
T(air) = 25°C
ΔT = T - 25°C
Q = 25 * 0.0968 * ( T - 25)
Q = 2.42 (T - 25)
Substitute Q = 48 into the equation above
48 = 2.42 (T - 25)
T - 25 = 19.84
T = 25 + 19.84
T = 44.84 ⁰C
Answer:
R=2F
Explanation:
As the forces are in same direction so the resultant force will be:
R=F+F
R=2F
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
