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Misha Larkins [42]

2 years ago

11

15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

the second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision

1 answer:

kolbaska11 [484]2 years ago

5 0

Answer:

Explanation:

Given that,

Mass of first car

M1= 328kg

The car is moving in positive direction of x axis with velocity

U1 = 19.1m/s

Velocity of second car

U2 = 13m/s, in the same direction as the first car..

Mass of second car

M2 = 790kg

Velocity of second car after collision

V2 = 15.1 m/s

Velocity of first car after collision

V1 =?

This is an elastic collision,

And using the conservation of momentum principle

Momentum before collision is equal to momentum after collision

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The velocity of the first car after collision is 14.04 m/s

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Answer:

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Explanation:

Power = Work / Time

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2 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

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Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago

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3 years ago

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hodyreva [135]

Answer:

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A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.

lyudmila [28]

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Current in the secondary, $I_{s} = 8.50\ mA$

Now,

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$N_{s}$ = No. of turns in secondary

$\frac{N_{s}}{N_{p}} = \frac{13000}{120}$ ≈ 108

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Power, P = $V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW$

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$\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$

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6 0

3 years ago