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Marta_Voda [28]

3 years ago

14

Train car A is at rest when it is hit by train car B. The two cars, which have the same mass, are stuck together and move off af

ter the collision. How does the final velocity of train cars A and B after the collision compare to the initial velocity of train car B before the collision? The final velocity is double train car B’s initial velocity. The final velocity is the same as train car B’s initial velocity. The final velocity is half of train car B’s initial velocity. The final velocity is zero since train car B will stop.

2 answers:

olga_2 [115]3 years ago

7 0

the answer is c. The final velocity is half of train car B’s initial velocity.

k0ka [10]3 years ago

4 0

The final velocity is half of train car B's initial velocity.

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When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

$2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}$

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

$2t = 2\lambda_{film}$

$t = \lambda_{film}$

The index of refraction of soap is given, then

$\lambda_{film} = \frac{\lambda_{vacuum}}{n}$

Combining the results of all steps we get

$t = \frac{\lambda_{vacuum}}{n}$

Rearranging, we find

$\lambda_{vacuum} = tn$

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4 0

3 years ago

A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm

____ [38]

Answer:

53.13 °

Explanation:

In order to do this, we just need to apply the following:

tanα = Dy/Dx

Where:

Vy: speed of the ball in the y axis.

Vx: speed of the ball in the x axis.

At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.

According to this, then:

tanα = (40/30)

tanα = 1.3333

α = tan⁻¹(1.3333)

<h2>α = 53.13°</h2>

This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.

Hope this helps

6 0

3 years ago

If you put two identical cars on opposite sides of a large magnet, what happens

FrozenT [24]

Answer:

Depends on what pole it is.

Explanation:

If the poles of the cars and magnets are the same they will repel, if different, attracts.

7 0

3 years ago

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A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

4 years ago

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serg [7]

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4 years ago

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