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Nataliya [291]
1 year 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]1 year 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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Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, t_{K} = 17.9 min = 17.9\times 60 = 1074 s

Time taken by Hannah, t_{H} = 15.3 min = 15.3\times 60 = 918 s

Now, the speed of Kara and Hannah can be calculated respectively as:

v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s

v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s

Time taken in each lap is given by:

(v_{H} - v_{K})t = x

(5.45 - 4.65)\times t = 400

t = \frac{400}{0.8}

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

d_{H} = v_{H}\times t

d_{H} = 5.45\times 500 = 2725 m

No. of laps taken by Hannah when she passes Kara:

n_{H} = \frac{d_{H}}{x}

n_{H} = \frac{2725}{400} = 6.8 ≈ 7 laps

3 0
1 year ago
A ball is dropped from the top of an eleven-story building to a balcony on the ninth floor. In which case is the change in the p
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Answer:

at the top of the 9 story building i think

Explanation:

5 0
2 years ago
Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288
Tomtit [17]
<span><u>Answer </u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football. 
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2 years ago
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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

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x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

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<span>The work is being done against gravity, and equals (Force * vertical distance).
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