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Katena32 [7]
3 years ago
5

A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. C

alculate the work done in J
Physics
1 answer:
Vlada [557]3 years ago
5 0

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

W=-P\Delta V=-P(V_2-V_1)

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm    (760torr =1atm)

V_1 = initial volume = 5.68 L

V_2 = final volume = 2.35  L

Putting values in above equation, we get:

W=-0.96atm\times (2.35-5.68)L=3.20L.atm

To convert this into joules, we use the conversion factor:

1L.atm=101.33J

So, 3.20L.atm=3.20\times 101.3=324J

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

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kykrilka [37]

Answer: Got It!

<em>Explanation:  </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

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13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

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3 years ago
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

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5 0
3 years ago
A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

-F_{f} + ma = 0      

\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

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3 years ago
Given the distance between the crest of one wave and the crest of the next wave, you can determine the?
Sonja [21]
Answer: wavelength !!
hope this helped :)
3 0
2 years ago
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