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Deffense [45]
2 years ago
6

engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi

on between two train cars traveling toward each other. Car one is traveling south at 18 M/S and has a full load, giving it a total mass of 14,650 kg. Kartoo is traveling north at 11 m/s and has a mass of 3825 kg. After the collision car one has a final velocity of 6 M/S south what is the final velocity of car 2?
Physics
2 answers:
IRISSAK [1]2 years ago
8 0

Answer:

35 m/s south

Explanation:

took the test

Archy [21]2 years ago
4 0

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

where ;

m_1=mass of object 1\\v_1i=initial  velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2

Given in the question that;

m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?

Apply the formula as;

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

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A man does 500 j work pushing a car a distance of 2m how much force does he apply
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Answer: 250 N

Explanation:

Use equation for work

W=F*d

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W=500J

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W=Fd

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3 years ago
A car has a mass of 8000 kg and a kinetic energy of 24,000 J. What is its<br> speed?
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8 0
3 years ago
In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c
shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

P = \frac{F}{A}

Hydraulic Lift - After change

P + \Delta P = \frac{F + \Delta F}{A}

Where:

P - Hydrostatic pressure, measured in pascals.

\Delta P - Change in hydrostatic pressure, measured in pascals.

A - Cross sectional area of the hydraulic lift, measured in square meters.

F - Hydrostatic force, measured in newtons.

\Delta F - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}

\Delta P = \frac{\Delta F}{A}

\Delta F = A\cdot \Delta P

Given that \Delta P = 100\,Pa and A = 25\,m^{2}, the additional weight is:

\Delta F = (25\,m^{2})\cdot (100\,Pa)

\Delta F = 2500\,N

The additional mass needed for the additional weight is:

\Delta m = \frac{\Delta F}{g}

Where:

\Delta F - Additional weight, measured in newtons.

\Delta m - Additional mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

If \Delta F = 2500\,N and g = 9.807\,\frac{m}{s^{2}}, then:

\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }

\Delta m = 254.92\,kg

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0
3 years ago
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