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Ymorist [56]
3 years ago
12

If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues tr

aveling at 25.0 m/s, with what velocity will the two move if they lock bumpers after a rear-end collision?
Physics
2 answers:
Alborosie3 years ago
5 0

Answer: 20.2 m/s

Explanation:

From the question above, we have the following data;

M1 = 800kg

M2 = 1200kg

V1 = 13m/s

V2 = 25m/s

U (common velocity) =?

M1V1 + M2V2 = (M1 + M2). U

(800*13) + (1200*25) = (800+1200) * U

10400 + 30000 = 2000u

40400 = 2000u

U = 40400 / 2000

U = 20.2 m/s

sashaice [31]3 years ago
3 0

Answer:

20.2 m/s

Explanation:

Using the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the sport car, m' = mass of the pick-up truck, u = Initial velocity of the sport car, u' = Initial velocity of the pick-up truck V = Common velocity.

make V the subject of the equation

V = (mu+m'u')/(m+m')............... Equation 2

Given: m = 800 kg, m' = 1200 kg, u = 13 m/s, u' = 25 m/s

Substitute into equation 2

V =[(800×13)+(1200×25)]/(800+1200)

V = (10400+30000)/2000

V = 40400/2000

V = 20.2 m/s

Hence both will move with a velocity of 20.2 m/s

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3 years ago
How to find the gradient of a velocity time graph
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The area-

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3 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
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Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

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to find out

period of oscillations

solution

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force = k × x   .........1

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3.43 = k × 0.12

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put here value

period of oscillations = 2π × \sqrt{\frac{0.35}{28.58} }  

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3 years ago
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<span>509=9.405ΔT </span>
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<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
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