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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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What hidden costs of using DDT did Rachel Carson call attention to?
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Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=
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Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2

v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2

Given that the initial velocities are equal in magnitude, we have;

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v₁ and v₂ = The final velocities of the satellites

We get;

v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095

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The final relative velocity of the satellite, v_f = v₁ + v₂

∴ v_f = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, v_f = 0.190 m/s

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