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Nataliya [291]
2 years ago
15

An open-topped freight car with mass 24,000 kg is coasting without friction along a level track. It is raining very hard, and th

e rain is falling vertically downward. Originally, the car is empty and moving with a speed of 4.00 m/s.(a) What is the speed of the car after it has collected 3000 kg of rainwater?(b) Since the rain is falling downward, how is it able to affect the horizontal motion of the car?
Physics
1 answer:
skelet666 [1.2K]2 years ago
7 0

Answer:

(a) v = 3..6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

Explanation:

from the question we have the following:

mass of the car (Mc) = 24,000 kg

initial velocity of the car (u) = 4 m/s

mass of water (Mw) = 3000 kg

final velocity of the car (v) = ?

(a) we can calculate the final momentum of the car by applying the conservation of momentum where

initial momentum = final momentum

Mc x U = (Mc + Mw) x V

24000 x 4 = (24000 + 3000) x v

96,000 = 27000v

v =3.6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

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Answer:

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Explanation:

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so we have

-\frac{d\phi}{dt} = EMF

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now we have

B = 50 sin(10\pi t)

A = \pi r^2 = \pi(0.04)^2 = 5.03 \times 10^{-3} m^2

now we have

EMF = -(40)(5.03 \times 10^{-3})\frac{d(50 sin(10\pi t))}{dt}

EMF = -0.2012(50 \times 10\pi)cos(10\pi t)

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now at t = 0.10 s

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3 0
2 years ago
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The work done on the barbell is -165.62 Nm.

Explanation:

Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.

In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.

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So, Work done = F*d

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At a point on the streamline, Bernoulli's equation is
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At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
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