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3 years ago

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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?

1 answer:

skelet666 [1.2K]3 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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Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

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3 years ago

Consider two tubes filled with water at the same height, one with fresh water and the other tube with salt water. The pressure i

Olegator [25]

Answer:

B

Explanation:

The correct option for the question is B that is salt water. In salt water, the density of water is higher so the pressure at the end of tube containing salt water will be greater. As according to the hydrostatic law the pressure at a given point will be directly proportional to the distance travelled as well.

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Oliga [24]

OPTION D is the correct answer.

Refer to the attachment for complete calculation...

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Learning Goal: To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following gene

kaheart [24]

Answer:

so the transverse displacement is 0.089963 m

Explanation:

Given data

equation y(x,t)=Acos(kx−ωt)

speed v = 9.00 m/s

amplitude A = 9.00 × 10^−2 m

wavelength λ = 0.480 m

to find out

the transverse displacement

solution

we know

v = angular frequency / wave number

and

wave number = 2 $\pi$ / λ = 2 $\pi$ / 0.480 = 13.0899 $m^{-2}$

angular frequency = v k

angular frequency = 9.00 × 13.0899

angular frequency = 117.8097 rad/sec = 118 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)

when x =0 and and t = 0

maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))

maximum y(x,t)= 9.00 × 10^−2 m

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y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )

y(x,t)=9.00 × 10^−2 × (0.99959)

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