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natta225 [31]
3 years ago
8

A chamber fitted with a piston can be controlled to keep the pressure in the chamber constant as the piston moves up and down to

increase or decrease the chamber volume. The chamber contains an ideal gas at 296 K and 1.00 atm. What is the work done on the gas as the piston compresses it from 1.00 L to 0.633 L
Physics
1 answer:
Troyanec [42]3 years ago
8 0

Answer:

The work done on the gas is 37 J.

Explanation:

let P be the pressure in the chamber.

let V1 be the initial volume of the chamber.

let V2 be the final volume of the chamber.

then, the work done on the gas is given by:

W = P(V2 - V1)

    = (1×10^5)(1 - 0.633)×10^-3

    = 37 J

Therefore, the work done on the gas is 37 J.

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Answer:

For the distance range 50 to 500 km, the S-waves travel about 3.45 km/s and the P-waves around 8 km/s.

hope it helps.

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In order to generate electricity, nuclear power plants take advantage of this part of the electromagnetic spectrum. gamma rays r
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Read 2 more answers
The speed of the object
serious [3.7K]
Speed of an object is kinetic energy
7 0
3 years ago
A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck trav
aniked [119]
<span>You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
 
Need to find xf, knowing xi = 0

Thus, use the formula
 x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
 x_f = 100m
 
so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

20m/s = xm/20s 20*20 = x
x = 400
 
thus we have 100+400 = 500m then it slows down from 500m to x_f
 
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5) x_f = 500 + 50
x_f = 550
 
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s

thus V_xavg = 16m/s</span>
8 0
3 years ago
An astronaut notices that a pendulum which took 2.45 s for a complete cycle of swing when the rocket was waiting on the launch p
klio [65]

Answer:

2.84 g's with the remaining 1 g coming from gravity (3.84 g's)

Explanation:

period of oscillation while waiting (T1) = 2.45 s

period of oscillation at liftoff (T2) = 1.25 s

period of a pendulum (T) =2π. \sqrt{\frac{L}{a} }

where

  • L = length
  • a = acceleration

therefore the ration of the periods while on ground and at take off will be

\frac{T1}{T2} =(2π \sqrt{\frac{L}{a1} } ) /  (2π\sqrt{\frac{L}{a2} })

where

  • a1 = acceleration on ground while waiting
  • a2 = acceleration during liftoff

\frac{T1}{T2} = \frac{\sqrt{\frac{L}{a1} }}{\sqrt{\frac{L}{a2} }}

squaring both sides we have

(\frac{T1}{T2})^{2} = \frac{\frac{L}{a1} }{\frac{L}{a2} }

(\frac{T1}{T2})^{2} = \frac{a2}{a1}

assuming that the acceleration on ground a1 = 9.8 m/s^{2}

(\frac{T1}{T2})^{2} = \frac{a2}{9.8}

a2 = 9.8 x (\frac{T1}{T2})^{2}

substituting the values of T1 and T2 into the above we have

a2 = 9.8 x (\frac{2.45}{1.25})^{2}

a2 = 9.8 x 3.84

take note that 1 g = 9.8 m/s^{2} therefore the above becomes

a2 = 3.84 g's

Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.

5 0
4 years ago
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