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For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
Answer:
v₃ = 9.62[m/s]
Explanation:
To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.
We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.
Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.
In this way, we can construct the following equation.

where:
m₁ = mass of the astronaut = 157 [kg]
m₂ = mass of the toolkit = 5 [kg]
v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]
v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]
v₃ = velocity at which the toolkit should be thrown [m/s]
Now replacing:
![-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]](https://tex.z-dn.net/?f=-%28157%2B5%29%2A0.2%3D%28157%2A0.1%29-%285%2Av_%7B3%7D%29%5C%5C%285%2Av_%7B3%7D%29%3D%2015.7%2B32.4%5C%5Cv_%7B3%7D%3D9.62%5Bm%2Fs%5D)
I think it's the last one
Answer:
The speed of sound is affected by temperature and humidity. Because it is less dense, sound passes through hot air faster than it passes through cold air. ... The attenuation of sound in air is affected by the relative humidity. Dry air absorbs far more acoustical energy than does moist air.