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Sauron [17]
3 years ago
6

Two car collide in an intersection. The speed limit in that zone is 30 mph. The car (mass of 1250 kg) was going 17.4 m/s (38.9).

The truck (2020 kg) t-boned the car in the middle of the intersection. The car was slowed down to only 6.7 m/s. The truck after colliding with the car was going 10.3 m/s. How fast did the truck go into the intersection?
Physics
1 answer:
Musya8 [376]3 years ago
5 0

Answer:

u₂ = 3.7 m/s

Explanation:

Here, we use the law of conservation of momentum, as follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\

where,

m₁ = mass of the car = 1250 kg

m₂ = mass of the truck = 2020 kg

u₁ = initial speed of the car before collision = 17.4 m/s

u₂ = initial speed of the tuck before collision = ?

v₁ = final speed of the car after collision = 6.7 m/s

v₂ = final speed of the truck after collision = 10.3 m/s

Therefore,

(1250\ kg)(17.4\ m/s)+(2020\ kg)(u_2)=(1250\ kg)(6.7\ m/s)+(2020\ kg)(10.3\ m/s)\\\\(2020\ kg)(u_2) = 8375\ N.s + 20806\ N.s - 21750\ N.s\\\\u_2=\frac{7431\ N.s}{2020\ kg}

<u>u₂ = 3.7 m/s</u>

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Answer:

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Explanation:

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P = absolute pressure

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the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

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W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

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dividing both equations

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replacing values

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Answer:

Explanation:

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We know that, 1rev = 2πrad

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Length of pedal r = 30cm = 0.3m

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