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garik1379 [7]
2 years ago
8

Beaker contain 200 mL of water what is volume of water in cm cube and m cube

Physics
2 answers:
Rainbow [258]2 years ago
7 0

1 mL = 1 cm³

1 cm³ = 0.000 001 m³

200 mL = 200 cm³

200 cm³ = 0.000 2 m³

It can be ANY substance. It doesn't have to be water.

maksim [4K]2 years ago
6 0

<u><em>Part 1:</em></u>

<u>The basics of conversions state that:</u>

1 ml is equivalent to 1 cm³

This means that the ratio between them is 1:1

Therefore, 200 ml of water would be equivalent to 200 cm³ of water

<u><em>Part 2:</em></u>

<u>Again, the basics of conversion states that:</u>

1 ml is equivalent to 1 * 10⁻⁶ m³

We will use cross multiplication to convert 200 ml to m³ as follows:

1 ml ................> 1 * 10⁻⁶ m³

200 ml .........> ??

200 ml = \frac{200*1*10^{-6}}{1}  = 0.0002 m³

Hope this helps :)

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rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

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2 years ago
If you and 3 friends are playing tug of war and are pulling with a force of 35 N and your teammate with 37 N, while your opponen
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Use appropriate units and significant figures.  USE THE LAW OF COSINES AND LAW OF SINES.
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Answer:

The resultant velocity is 86.1 mi/h.    

Explanation:

The law of cosines is given by:

c^{2} = a^{2} + b^{2} - 2abcos(\theta)

Where:

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a: is the velocity of the plane = 75.0 mi/h

b: is the velocity of the wind = 15.0 mi/h  

θ: is the angle between "a" and "b"                          

The angle between "a" and "b" can be found as follows:

\theta = 180.0 - 46.0 = 134.0 ^{\circ}

Now, by using the law of cosines we have:

c^{2} = (75.0)^{2} + (15.0)^{2} - 2*75.0*15.0*cos(134.0) = 7413.0

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Therefore, the resultant velocity is 86.1 mi/h.    

The law of sines is:

\frac{a}{sin(\gamma)} = \frac{b}{sin(\alpha)} = \frac{c}{sin(\theta)}

Where:

γ: is the angle between "b" and "c"

α: is the angle between "a" and "c"

So, if we want to find "c" by using the law of sines, we need to know another angle besides θ (γ or α), and the statement does not give us.

I hope it helps you!        

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