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Yuri [45]
3 years ago
14

A railroad tank car contains milk and rolls at a constant speed along a level track. The milk begins to leak out the bottom. The

car then
a. Need more information about the rate of the leak.
b. maintains a constant speed.
c. speeds up.
d. slows down.
Physics
1 answer:
kifflom [539]3 years ago
8 0

Answer:

option C

Explanation:

The correct answer is option C                        

There is no external force acting in the system hence the momentum will be conserved.                                    

As the milk is leaking out of the tank mass of the tanker is decreasing.

When the mass of the container will decrease to  conservation the momentum speed of the container will have to be increased.

So, the car carrying milk will speed up.

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Size relationships between parts of a whole are known as.
Stella [2.4K]

Answer:

proportion

Explanation:

hope it was helpful to you

then

plz mark me as brainlist

8 0
2 years ago
A 1.60 m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the
Viefleur [7K]

Answer:

Explanation:

Given

mass of book(m)=2.1 kg

height up to which book is lifted is (h)2.2 m

height of person (h_0)1.6 m

Potential energy of book relative to ground=mgh

PE=2.1\times 9.8\times 2.2=45.276 J

(b)PE w.r.t to person head =mg(h-h0)

=2.1\times 9.8\times (2.2-1.6)=12.348 J

work done by person in lifting box 2.2 m w.r.t floor

Word done =Potential Energy of box relative to floor=45.2 J

6 0
3 years ago
KINDLYY FASTT A uniform metre rule of mass 100 g is pivoted at the 60 cm mark. At what point on the meter rule should a mass of
MakcuM [25]

Answer:

80 cm.

Explanation:

Since the metre rule is 1 m i.e 100 cm it means that the mass of the metre can be obtained at the 50 cm mark.

But the metre is pivoted at the 60 cm.

Please refer to the attached photo for details.

In the attached photo, y is the distance between the mark (that will balance the metre rule when a 50 g mass is hunged) and the pivot.

Thus, we can obtain the value of y as follow:

Anticlock wise moment = Clock wise moment

Anticlock wise moment = 100 × 10

Clock wise moment = y × 50

Anticlock wise moment = Clock wise moment

100 × 10 = y × 50

Divide both side by 50

y = (100 × 10) /50

y = 20 cm

Now, to obtain the mark, B that will balance the metre, we simply add 20 cm to 60 cm ie

B = 60 + 20

B = 80 cm

6 0
3 years ago
How can i pass physics exam in 3 days?​
alexira [117]

Explanation:

you can pass the exam in 3 days by doing labouring hard, by understanding and by thinking theoretical and practising practically.

5 0
3 years ago
The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much
aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}

Upon substituting our given values, we will get:

\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}

\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365

Taking square root of both sides, we will get:

\frac{T_1}{T_2}=\sqrt{0.0073878361246365}

\frac{T_1}{T_2}=0.0859525225030452495

\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}

\frac{T_2}{T_1}=11.634329870476699\approx 11.634

T_2=11.634\cdot T_1

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0
3 years ago
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