Physics, 17.01.2020 13:31 carlosatniel47

You see your friend 100 m away down the street. if you yell back and forth, your conversation will be affected by the speed of sound, which is 340 m/s. if, instead, you use your cell phone to call her on her cell phone, the conversation is relayed via a communications satellite that is 36000 km above you. what is the delay between a sound leaving your mouth and its arrival at your friend’s ear if (a) you yell, or (b) you use your cell phone? (c) how far would you have to be from your friend for the delays to be equal?

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Answer from: fatamhassan21

(a) 0.29 s

If you yell, your voice will reach your friend by travelling as a sound wave through the air.

The speed of sound in air is

v = 340 m/s

while the distance to be covered is

d = 100 m

So, the time taken is

$t=\frac{d}{v}=\frac{100 m}{340 m/s}=0.29 s$

(b) 0.24 s

In this case, the voice is transmitted as a radio wave (electromagnetic wave) to the satellite and then back.

The speed of electromagnetic waves is the speed of light:

$c=3\cdot 10^8 m/s$

while the distance the wave has to cover is twice the distance between the ground and the satellite:

$d=2 \cdot 36000 km=72000 km=7.2 \cdot 10^7 m$

So, the time taken is

$t=\frac{d}{c}=\frac{7.2\cdot 10^7 m}{3\cdot 10^8 m/s}=0.24 s$

In order for the two delays to be equal, the distance between you and your friend (d') must satisfy the equation

$\frac{d'}{v}=\frac{d}{c}$

where on the left is the time taken by the sound to travel through the air, while on the right is the time taken by the radio wave to travel back and forth from the satellite.

Solving the equation for d',

$d'=v\frac{d}{c}=(340 m/s)\frac{7.2\cdot 10^7 m}{3\cdot 10^8 m/s}=81.6 m$

Answer from: Quest

Ithink it’s 98 because jajajjajansmsmjs

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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