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3 years ago

7

A balloon of hydrogen is put into to pressure chamber. The initial pressure and volume of hydrogen is 1 atm and .5 cm3. The pres

sure of the chamber is increased to 2 atm. What is the new volume?

1 answer:

Elis [28]3 years ago

4 0

Answer:

0.25 cm³.

Explanation:

We shall apply Boyle's law to find the solution . According to it

PV = constant where P is pressure and V is volume of the gas.

P₁ V₁ = P₂V₂

1 x .5 = 2 x V₂

V₂ = 0.25 cm³.

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Write a hypothesis about the effect of the wavelength on the diffraction angle. Use the "if . . . then . . . because . . ." form

adell [148]

A) Wavelength is the distance between two successful crests or troughs in a transverse wave or the distance between two successful compressions or rarefactions in a longitudinal wave. if the wavelength of a wave is long then the wave will diffracts more compared to a wave with a shorter wavelength because the size of the wavelength is proportional to the angle of diffraction. Such that if the gap is larger than the wavelength then the wave passes through the gap and does not spread much on the other side, while when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread greatly out.

b) Diffraction is a phenomenon that occurs when a wave encounters an obstacles or a slit. It is referred to as the bending of light around corners of an obstacles or aperture into the region of geometrical shadow of the obstacle. It is caused by one wave of light is shifted by the diffracting object which will in turn cause the wave to have interference with itself. (either constructive or destructive)

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3 years ago

Read 2 more answers

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

In this case:

$cos \theta = \frac{r}{R_{E}}$

so we can solve for r, so we get:

$r= R_{E}cos \theta$

in this case we'll use the average radius of earch which is 6,371 km, so we get:

$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

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3 years ago

Is there any difference between reflection and echo???

Fofino [41]

Answer:

reflection is the act of reflecting or the state of being reflected while echo is

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3 years ago

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laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

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When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

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4 years ago

State newtons third law of motion,and give an example that shows how this law works

julsineya [31]

his third law is for every action, there is an equal and opposite reaction. For example when you throw a ball on the floor, it bounces back up. The action is throwing the ball and the reaction is the ball bouncing back up.

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3 years ago