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____ [38]
3 years ago
9

Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the

spectator A is 0.55 s, and for the spectator B it is 0.45 s. Sight lines from the two spectators to the player kicking the ball meet at an angle of 90°. The speed of sound in the air is 343 m/s.
How far are (a) spectator A and (b) spectator B from the player?
(c) How far are the spectators from each other?
Physics
1 answer:
WINSTONCH [101]3 years ago
5 0

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

t_A=0.55s

t_B=0.45s

(a) The distance from the kicker to each of the 2 spectators is given by:

d_A=v \times t_A

where,

v= speed of sound

t_A=time taken for the sound waves to reach the ears

d_A=343\times 0.55=188.65m

(b)d_B=v \times t_B

where,

v= speed of sound

t_B=time taken for the sound waves to reach the ears

d_B=343\times 0.45=154.35m

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m

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\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm

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If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 . (a) what
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