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ikadub [295]
3 years ago
6

A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the

same momentum as the other cars collides with an enormously massive wall. Which of the three cars will experience the greatest impulse
Physics
2 answers:
WARRIOR [948]3 years ago
5 0

Answer:

The head on Collison because you have both cars going (for the sake of it 30mph) and they both collide the energy from that is 60mph because the speed is combined with the two cars.

Marat540 [252]3 years ago
4 0

All three cars experience the same impulse.

Impulse is equal to change in momentum.

Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.

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Mariulka [41]

Answer:

=3.5 m/s

Explanation:

y = x tanθ - 1/2 g x² / (u²cos²θ )

y = 0.25 , x = 0.5, θ = 40°

.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40

.25 = .42 - 2.0875/u²

u = 3.5 m / s.

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A party shop delivers helium-filled balloons to homes and businesses. The owners realize from experience that on hot summer days
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6 0
2 years ago
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What happens in the gray zone between solid and liquid?
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Answer:

Hence from liquid to solid or solid to liquid the transition has to cross the grey zone. This grey zone transition is is very crucial which includes the intermolecular forces acting on the molecules and each atoms which makes the change in state from hot to cold and cold to hot.

Explanation:

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3 years ago
Using the results of Question 1 that would apply if the collision were inelastic, compute (using Excel in the yellow highlighted
ch4aika [34]

Answer:

The fractional kinetic energy will be lost if the collision is inelastic. In inelastic collision, the kinetic energy is converted into other forms of energy.

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Explanation:

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4 0
3 years ago
A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

fe =  focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

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m = m_{o}* m_{e}

m = 14*26 = 364

4 0
3 years ago
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