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jeyben [28]
3 years ago
8

You are a working for a rail company and are studying the effects of rain on open train cars. As a test case, you consider a 6,3

25 kg open train car rolling on frictionless rails at 27.4 m/s when it starts pouring rain. If 1,250 kg of rain fall into the train car over a few minutes, what will the car's speed become?
Physics
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

<em>22.87 m/s</em>

<em></em>

Explanation:

mass of the train car = 6325 kg

velocity of the train = 27.4 m/s

mass of rain that falls into the train car = 1250 kg

the train car's final speed = ?

Initially, the train has a momentum of

momentum = (mass of train car) x (velocity of the train car)

momentum = 6325 x 27.4 = 173305 kg-m/s

When the mass of water falls into the train, the mass of the system increase to include that of the mass of the rain, but the momentum of the system remains the same according to the laws of conservation of linear momentum.

The new mas = (mass of train car) + (mass of rain)

==> 6325 + 1250 = 7575 kg

The final velocity of the system can be found by using the relationship

initial momentum = final momentum

Final momentum of the system = (final mass of the system) x (final velocity of the system)

==> 7575 x v = 7575v

where v is the final mass of the train car-rain system

Equating the final and initial momenta together, we have

173305 = 7575v

v = 173305/7575 = <em>22.87 m/s</em>

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leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have

<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>

where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

3 0
3 years ago
Can a compound be separated by physical means?
Julli [10]

No,  it is not possible.
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Any elements that have been joined together chemically can only be separated back into their constituent elements by chemical means because the bonds holding them together can only be broken using chemical means.

A good example is sodium chloride, table salt. Poisonous chlorine gas and toxic sodium metal react together whereby sodium loses one electron which chlorine readily accepts and in the process an ionic bond is formed between the two resulting in a totally new, harmless compound , sodium chloride.

Only through electrolysis can sodium chloride be separated back into sodium and chlorine gas. No physical means can be used to do that.

3 0
3 years ago
The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will b
pogonyaev

Answer:

Lower

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gsintheta (gsinθ)

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h = heigth = 110 meters

Replacing:

PE = 3800 * 10 * 110 = 4,180,000 J

7 0
1 year ago
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