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2 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]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:

Time, t = 0.23 seconds

Explanation:

It is given that,

Initial speed of the ranger, u = 52 km/h = 14.44 m/s

Final speed of the ranger, v = 0 (as brakes are applied)

Acceleration of the ranger, $a=-4\ m/s^2$

Distance between deer and the vehicle, d = 87 m

Let d' is the distance covered by the deer so that it comes top rest. So,

$d'=\dfrac{v^2-u^2}{2a}$

$d'=\dfrac{-(14.44)^2}{2\times -4}$

d' = 26.06 m

Distance between the point where the deer stops and the vehicle is :

D=d-d'

D=87 - 26.06 = 60.94 m

Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :

$t=\dfrac{v}{D}$

$t=\dfrac{14.44}{60.94}$

t = 0.23 seconds

Hence, this is the required solution.

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a 20 ft shipping container on a cargo ship has a mass of 24000 kg and a volume of 33.2m3. what is the density of the shipping co

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You are at the edge of a diving board that is 9 meters above the water. If you weigh 500 Newtons, what is your potential energy?

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Answer:

4500 J

Explanation:

First, let's define some equations and derivations.

Our potential energy formula is:

$\displaystyle U = mgh$

Where <em>m </em>is mass (in kg), <em>g</em> is the gravitational constant (in m/s²), and <em>h</em> is height (in m).

We also know that <em>mg</em> is equal to the weight of an object (in N), from Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration).

Therefore, we can simply substitute force into the equation:

$\displaystyle U = Fh$

Where <em>F</em> is the force (in N) and <em>h</em> is still height (in m).

Now we can calculate the amount of potential energy in our system, measured in joules.

Substitute in the given variables, F = 500 N and h = 9 m:

$\displaystyle U = (500 \ N)(9 \ m)$

Using simple Pre-Algebra rules, we find that:

$\displaystyle U = 4500 \ J$

This tells us that the we have 4500 joules of potential energy when I am 9 meters above the water on the edge of the diving board.

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Answer:

you exert an Applied Force on an object when you pull it towards you

A. Applied Force

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