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gizmo_the_mogwai [7]
4 years ago
14

A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

the temperature is 304 K.
Physics
1 answer:
iogann1982 [59]4 years ago
7 0

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

p_2= 1.76 atm

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Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

v_A = 35 mi/h

and the distance travelled is

d = 10 mi

so the time taken for car A is

t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h

The speed of car B is

v_B = 45 mi/h

and the distance travelled is

d = 10 mi

so the time taken for car B is

t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h

So the difference in time is

\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h

Which corresponds to

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so car B arrived 230.4 s before car A.

5 0
3 years ago
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
4 years ago
What are the necessary condition for the production of sound
andre [41]

Explanation:

You need two, maybe three things - something that's vibrating, a medium for those vibrations to propagate in, and a listener to hear it or recording equipment to pick it up

8 0
3 years ago
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8 0
3 years ago
The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons
bulgar [2K]

Answer:\lambda =248.99 nm

Explanation:

Given

Work function\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98

h=6.626\times 10^{-34} J

c=2.998\times 10^8

\phi =\frac{hc}{\lambda }

\lambda =\frac{hc}{\phi }

\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}}

\lambda =248.99 nm

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4 years ago
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