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Artyom0805 [142]

2 years ago

9

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i

s kept constant?"

1 answer:

asambeis [7]2 years ago

5 0

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

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A boat moves through the sea.

sergiy2304 [10]

Answer:

dont you have to times it

Explanation:

4 0

2 years ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

The initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This energy of proton is $\simeq 250 GeV$

Thus the speed of the proton, v $\simeq c$

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

Now, in accordance to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0

2 years ago

Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.

solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination $\theta =58.7^{\circ}C$

Horizontal distance travel by Particle $d=1200 m$

Vertical height $h=780 m$

Let u be the initial velocity

calculating vertical distance

$y=u\sin \theta +\frac{at^2}{2}$

$y=u\sin \theta t-\frac{gt^2}{2}$ -------1

Calculating horizontal distance

$x=u\cos \theta \times t+0$

$t=\frac{x}{u\cos \theta }$

put value of t in equation 1

$y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2$

$y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }$

$\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y$

$u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}$

$u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}$

at $y=-780\ m\ x=1200 m$

$u^2=\frac{18.154}{2753.65}\times 1200^2$

$u=97.41 m/s$

6 0

2 years ago

Calculate the binding energy e of the boron nucleus 11 5b (1ev=1.602×10−19j). express your answer in millions of electron volts

nignag [31]

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7 0

2 years ago