Answer:
The aeroplane flew 4200 miles for 7 hours.
600 times 7 is 4200.
Answer:
8.854 pF
Explanation:
side of plate = 0.1 m ,
d = 1 cm = 0.01 m,
V = 5 kV = 5000 V
V' = 1 kV = 1000 V
Let K be the dielectric constant.
So, V' = V / K
K = V / V' = 5000 / 1000 = 5
C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F
C = 8.854 pF
Answer:
0.08 ft/min
Explanation:
To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.
So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:
since the difference between the upper and lower base is the increase in the base and we are only at halft the height.
Now we can calculate the longitudinal section <em>A</em> at that point:
And the raising speed <em>v </em>of the water is given by:
where <em>q</em> is the water flow (1 cubic foot per minute).
Answer:
ωf = 0.16 rad/s
Explanation:
Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²
Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²
(ASSUMING it is a uniform disk)
Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s
Conservation of angular momentum
44.8 = (51.2 + 230.4)ωf
ωf = 0.15909090...
Answer:
2 m/s
Explanation: Given that a flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe) with a velocity of 6 m/s to the west.
Since the jogger is moving in an opposite direction to the direction of the train, and velocity is the distance covered in a specific direction, the jogger will be moving at a velocity relative to the velocity of the train.
Velocity = (8 - 6) m/s
Velocity = 2 m/s
Therefore, the jogger will be moving at the speed of 2 m/s
