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bearhunter [10]

3 years ago

5

An object floats on water with 80% of its volume below the surface. the same object when placed in another liquid floats on that

liquid with 72% of its volume below the surface. determine the density of the unknown fluid.

1 answer:

GarryVolchara [31]3 years ago

8 0

0.9g/cm cubed, this is my full answer however i have to write at least 20 characters so this is filler :)

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

Which describes sublimation?

balu736 [363]

A. Solid turns into a gas

5 0

3 years ago

Hello help me pls! i need serious help

9966 [12]

Answer:c

Explanation:

7 0

3 years ago

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Two piles of heavy boxes must be lifted onto a shelf. Each pile has 20 boxes and each box is the same size and weight. Tom and c

Murrr4er [49]

Tom used more Force but over a shorter distance. Tom and Claudia both did the same amount of work.

4 0

4 years ago

Read 2 more answers

Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the

nirvana33 [79]

Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

$S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }$

where

$S_A$ indicates the speed of the ship A

$S_B$ indicates the speed of the ship B

and

C indicates the velocity of life

now we will Substitute 0.40c for A and 0.60 for B in the equation which is

$S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }$

after solving the above equation we will get

0.80 c

So, The correct answer is 0.80c

5 0

3 years ago