Answer:
<em>Before the parachute opens: Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.</em>
<em>Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.</em>
<h3>
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Answer:
r = 5.07 m
Explanation:
given,
velocity of the man , v = 3.43 m/s
centripetal acceleration, a = 2.32 m/s²
magnitude of position of = ?
using centripetal acceleration formula



r = 5.07 m
The magnitude of the position vector relative to rotational axis is equal to 5.07 m.
Answer:
168 seconds (2 min 48 s)
Explanation:
Find the heat absorbed by the water.
q = mCΔT
q = (1 kg) (4200 J/kg/K) (70°C − 40°C)
q = 126,000 J
Power is energy per time.
P = q / t
750 W = 126,000 J / t
t = 168 s
It takes 168 seconds (2 min 48 s).
Answer:
C
Explanation:
It is the answer I think let me know
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