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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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olga2289 [7]

The force result in stretching the spring 10.0 centimeters is 2.5N.

<h3>What is Hooke's law?</h3>

If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

F = kx

where k is the proportionality constant called the spring constant or force constant.

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.

Let the spring constant be very low 0.04N/m

The force applied is

F = 10 cm / 0.04

F = 0.1 m  / 0.04

F = 2.5 N

Thus, the force result in stretching the spring 10cm is 2.5 N.

Learn more about hooke's law.

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5 0
1 year ago
The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25
Vikki [24]

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

6 0
3 years ago
2. a) Force affects the body and results in different changes. The grid given below
postnew [5]

Answer:

Please find the answer in the explanation

Explanation:

Given that Force affects the body and results in different changes.

The grid that has the words related to the effects of force are :

a) Force could be a push or a pull.

c) Force has magnitude as well as direction.

e) Force acting on an object may cause a change in its state of motion or a change in its shape.

f) A force can act on an object with or without being in contact with it.

8 0
3 years ago
This is a green pigment in chloroplasts that traps light energy from the sun.
vovikov84 [41]
green pigment in the chloroplast is chlorophyll
6 0
3 years ago
"Suppose you tie a rock to the end of a 0.96 m long string and spin it in a horizontal circle with a constant angular velocity o
Ber [7]

Answer:

The mass of the rock is  m = 2.46406 \ kg

Explanation:

From the question we are told that

   The length of the string is  l  = 0.96 \ m

    The angular velocity is  w =  20.25 \ rad/s

     The tension on the string is T  = 970 \ N

Generally the centripetal force acting on the rock is mathematically evaluated as

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making m the subject of the formula

      m = \frac{T}{w^2 * l}

substituting values

     m = \frac{970}{(20.25^2) * (0.96)}

     m = 2.46406 \ kg

       

8 0
2 years ago
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