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Morgarella [4.7K]
3 years ago
7

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

e time delay for spectator A is 0.231 s and for spectator B is 0.128 s. Sight lines from the two spectators to the player kicking the ball meet at an angle of 90°. How far are (a) spectator A and (b) spectator B from the player? (c) How far are the spectators from each other? (Take the speed of sound to be 343 m/s.)

Physics
1 answer:
Akimi4 [234]3 years ago
6 0

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

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h = 23.716 m

Explanation:

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The distance that the body will fall through the time is given by the formula

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Substituting the values in the above equation

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A student is given three wires that are made from different materials, but each wire has the same physical dimensions. For a giv
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Use the ammeter to measure the current that flows through each wire, because a larger current that flows through the wire corresponds to a smaller resistivity

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Question 1 of 10
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Answer:

C. 5.6 × 10^11 N/C

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The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

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Now, in our case

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