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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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vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

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According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

\sum F_x = ma_x

where

m is the mass of the block

a_x is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

a_x = 0

So the equation becomes

\sum F_x = 0 (1)

The net force here is given by

\sum F_x = F cos \theta - F_f (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N

Learn more about  force of friction:

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3 years ago
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Aleksandr [31]
I think 1 and 3 is absolutely right but im not sure about number 2.
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kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

                                                                          = 1092 ft³

The weight of the air = density  x  volume

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