Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direc
1 answer:
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Answer:
Magnitude of avg velocity,
Given:
Constant speed of train, v = 79 km/h
Time taken in East direction, t = 27 min =
Angle,
Time taken in east of due North direction, t' = 29 min =
Time taken in west direction, t'' = 37 min =
Solution:
Now, the displacement, 's' in east direction is given by:
Displacement in east of due North for 29.0 min is given by:
Now, displacement in the west direction for 37 min:
Now, the overall displacement,
(a) Now, average velocity, is given:
Magnitude of avg velocity is given by:
(b) angle can be calculated as:
Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9
That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor