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3 years ago

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At a pressure 42 kPa, the gas in a cylinder has a volume of 11 liters. Assuming temperature remains the same, if the volume of t

he gas is decreased to 9 liters, what is the new pressure?

1 answer:

nlexa [21]3 years ago

5 0

Boyle's law<span> is a gas </span>law<span>, stating that the pressure and volume of a gas have an inverse relationship , when temperature is held constant. That is PV = constant. Therefore, (PV)initial = (PV)final. 42x11 = 9x P(final). P(final) = 42x11/9 = 51.34kPa. </span>

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A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located

const2013 [10]

Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

$v = \lambda f$

here we know that the wavelength of the wave is

$\lambda = 2.5 m$

$f = 10 Hz$

now speed of the wave is given as

$v = 10(2.5)$

$v = 25 m/s$

Now time taken by the wave to reach 5 m distance is

$t = \frac{L}{v}$

$t = \frac{5}{25}$

$t = 0.2 s$

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3 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

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3 years ago

Explain how evaporation cools a liquid

MAVERICK [17]

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Hope that this was helpful :)

8 0

2 years ago

The momentum of an object is determined to be 7.2 x 10-3 cm kg x m/s. Express this as provided or use any equivalent unit. How i

Leno4ka [110]

Complete Question:

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (Note: 1 kg = 1000 g).

Answer:

7.2 gm/s.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

Given the following data;

Momentum = 7.2 * 10^-3 kgm/s

1 kg = 1000 g

Substituting the unit in kilograms with grams, we have;

Momentum = 7.2 * 10^-3 * 1000 gm/s

<em>Momentum = 7.2 gm/s. </em>

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3 years ago

A student walks 100 m East and 100 m North. What is their distance traveled?

vichka [17]

-- The student's distance traveled is 200 meters.

-- The student's displacement is 141 meters Northeast.

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3 years ago