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frosja888 [35]
3 years ago
6

1. A truck with a mass of 8, 000 kg is traveling at 26.8 m/s when it hits the brakes. A.)What is the momentum of the truck befor

e it hits the brakes?
B.)The truck comes to a stop, losing all of that momentum. It takes 12.4 s for the truck to come to a
complete stop. How much force does it take to stop the truck?
Physics
1 answer:
NikAS [45]3 years ago
8 0

Answer:

1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s

B.) The force it takes to stop the truck is approximately 17,290.4 N

Explanation:

1. A.) The given parameters are;

The mass of the truck, m = 8,000 kg

The velocity of the truck when it hits the brakes, u = 26.8 m/s

Momentum = Mass × Velocity

The moment of the truck = The mass of the truck × The velocity of the truck

Therefore;

The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s

B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck

The time it takes the truck to come to a complete stop, t = 12.4 s

The deceleration, "a" of the truck is given by the following kinematic equation of motion

v = u - a·t

Where;

v = The final velocity of the truck = 0 m/s

u = The initial velocity = 26.8 m/s

a = the deceleration of the truck

t = The time of deceleration of the truck = 12.4 s

Substituting the known values gives;

0 = 26.8 - a × 12.4

Therefore;

26.8 = a × 12.4

a = 26.8/12.4 ≈ 2.1613

The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²

Force = Mass × Acceleration

The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck

∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.

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Answer:

charge C = greatest net force

charge B = the smallest net force

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F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

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Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

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