As per the question, the velocity of the airplane [v] = 660 miles per hour.
The total time taken by airplane [t] = 3.5 hours.
We are asked to determine the total distance travelled by the airplane during that period.
The distance covered [ S] by a body is the product of velocity with the time.
Mathematically distance covered = velocity × total time
S = v × t
= 660 miles/hour ×3.5 hours
= 2310 miles.
Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.
Answer:
r = 58.44 [m]
Explanation:
To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.
a = v²/r
where:
a = centripetal acceleration = 15.4 [m/s²]
v = tangential speed = 30 [m/s]
r = radius or distance [m]
r = v²/a
r = 30²/15.4
r = 58.44 [m]
Answer:
-5.8868501529 m/s² or -5.8868501529g
0.118909090909 s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Dividing by g
The acceleration is -5.8868501529 m/s² or -5.8868501529g
The time taken is 0.118909090909 s
Answer:
doubled the initial value
Explanation:
Let the area of plates be A and the separation between them is d.
Let V be the potential difference of the battery.
The energy stored in the capacitor is given by
U = Q^2/2C ...(1)
Now the battery is disconnected, it means the charge is constant.
the separation between the plates is doubled.
The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.
C' = C/2
the new energy stored
U' = Q^2 / 2C'
U' = Q^2/C = 2 U
The energy stored in the capacitor is doubled the initial amount.