Answer:
Height with sound ignored = Gravity x Time taken = 9.8 x 8.60 = 84.28 meters
Time taken by the sound = 84.28/330 = 0.255 seconds
Height with sound involved = (84.28 x 0.255) + 84.28 = 105.80 meters
a. Underestimated
Answer:
1. Speed=0
2. 2.46 s
3.30.1 m
4. 22.0 m
5.1.004 s
Explanation:
We are given that
Initial speed of blue ball, u=24.1 m/s
Height of blue ball from ground y_0=0.5 m
Initial speed of red ball , u'=7.2 m/s
Height of red from ground=y'0=32 m
Gravity, g=
1.When the ball reaches its maximum height then the speed of the blue ball is zero.
2.v=0
Using the formula and substitute the values
Where g is negative because motion of ball is against gravity
3.
Using the formula
4.Time of flight for red ball=3.77-2.9=0.87s
Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.
5.According to question
Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.
Answer:
A) mr = 100 kg
B) Fr = 210N
C) Ft = -199.5N
Explanation:
By conservation of the momentum:
mt*Vo = (mr + mt) * Vf Solving for mr:
mr = mt*Vo / Vf - mt = 100 kg
The average force on the receiver:
mr *(Vf - 0) = Fr * Δt Solving for Fr:
Fr = 210 N
The average force on the tackler:
mt * (Vf - Vo) = Ft * Δt Solving for Ft:
Ft = -199.5 N
Answer:
The upstroke of the piston draws water, through a valve, into the lower part of the cylinder. On the downstroke, water passes through valves set in the piston into the upper part of the cylinder. On the next upstroke, water is discharged from the upper part of the cylinder via a spout.
Explanation:
Explanation:
The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.
Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.
∑τ = Iα
W (0.25 m) − (100 N) (0.50 m) = 0
W = 200 N
Sum the forces in the y direction.
∑F = ma
F − 100 N − 200 N = 0
F = 300 N
