Answer: a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C
Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference
As we know the capacity is value that depend on the geometry of the capacitor, in our case two concentric spheres.
So Potential Difference between the spheres is given by:
ΔV=-
Where E = k*Q/ r^2
so we have 
then
Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)
Finally using C=Q/ΔV=ab/(k(b-a))
To caclulate the electric firld we first obtain the charge
Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C
so E=KQ/r^2 for both values of r
r=12.8 cm ( in meters)
r2=14.7 cm
E(r1)=4.83* 10^3 N/C
E(r2)=3.66 *10^3 N/C
S(travel distance)=85 ft
v (velocity)=15 ft/s
-----------------------------------
t (time)=?
Calculate the time with the formula for the velocity:
v=S/t
t=S/v
t=85 ft/(15 ft/s)
t=5.666s
Answer:
The maximum power density in the reactor is 37.562 KW/L.
Explanation:
Given that,
Height = 10 ft = 3.048 m
Diameter = 10 ft = 3.048 m
Flux = 1.5
Power = 835 MW
We need to calculate the volume of cylinder
Using formula of volume
Put the value into the formula
We need to calculate the maximum power density in the reactor
Using formula of power density
Where, P = power density
E = energy
V = volume
Put the value into the formula
Hence, The maximum power density in the reactor is 37.562 KW/L.
Answer:
t = 12s
Explanation:
Given:
v-initial = 0 m/s
x = 360 m
a = 5.0 m/s^2
Solve:
x = (v-initial)t + 1/2(a*t^2)
360 = 0t + 1/2 (5.0t^2)
360 = 2.5 t^2
144 = t^2
t = sqrt(144) = 12
Therefore, it takes 12 seconds.
Answer:
A. The bomb will take <em>17.5 seconds </em>to hit the ground
B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it
Explanation:
Maverick and Goose are flying at an initial height of 
, and their speed is v=688 m/s
When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement
The equation for the height y with respect to ground in a horizontal movement (no friction) is
[1]
With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released
The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time
The range (horizontal displacement) of the bomb x is
[2]
Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:
Setting y=0 and isolating t we get
Since we have
Replacing in [2]
A. The bomb will take 17.5 seconds to hit the ground
B. The bomb will land 12040 meters on the ground ahead from where they released it
