3 years ago

Can you see gas in a bottle

Physics

2 answers:

Galina-37 [17]3 years ago

6 0

Depends on what type of gass

Gelneren [198K]3 years ago

4 0

I think you sometime can but most of the time can't. Most gases are in color containers. That makes it kind of hard to answer this question. I hope this helps!

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Two soccer players kick the same 1-kg ball at the same time in opposite directions. One kicks with a force of 25 N; the other ki

andre [41]

Answer:

The forces offset to produce a net force of 8 N

since the ball is 1 kg

F = ma

a = f/m = 8 N/1 kg = 8 (kg m)/(s^2 kg) = 8 m/s^2

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2 years ago

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Reflection off of a smooth surface like a mirror is an example of diffuse reflection.

Akimi4 [234]

Answer:

true

Explanation:

Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. ... This is called diffuse reflection. Diffuse reflection is when light hits an object and reflects in lots of different directions.

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3 years ago

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Ksivusya [100]

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2 years ago

A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r

bazaltina [42]

Answer:

$V=16.65 m^3$

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

$R=\frac{25}{100}*74kg=18.5kg$

So the net force uproad on the balloon is

$F_b=18.5kg*g$

Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

$P_h=\frac{m}{V}=0.179 kg/m^3$

$P_A=1.29 kg/m^3$

$F_b=F_A-F_H$

$F_b=m_a*g-m_h*g$

$m=P/V$

$18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g$

$V=\frac{18.5kg}{(1.29-0.179)kg/m^3}$

$V=16.65 m^3$

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2 years ago

An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a

Liono4ka [1.6K]

Answer:

$16.25^{\circ}$

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Horizontal range is given by

$R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}$

The angle at which the arrow is to be released is $16.25^{\circ}$ .

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2 years ago