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Paha777 [63]
3 years ago
12

One mole of ideal gas is slowly compressed to one-third of its original volume. in this compression, the work done on the gas ha

s magnitude 672 j . for the gas, cp=7r/2.
Physics
1 answer:
beks73 [17]3 years ago
4 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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What happens when someone is walking across the carpeted room then touches a metal door knob
igomit [66]
Static electricity travels to the door knob because of the friction caused by the feet on the carpet. the friction traveled through the person, to their hand, and to the door knob because it is the best conductor.
6 0
3 years ago
A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at
Gnom [1K]

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

W=p(V_f -V_i)

where in this case we have:

p = 2 bar = 2\cdot 10^5 Pa is the pressure

V_i = 0.1 m^3 is the initial volume

V_f = 0.12 m^3 is the final volume

Substituting,

W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ

The 1st law of thermodynamics also states that

\Delta U = Q-W

where

\Delta U is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

\Delta U = +0.25 kJ

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ

7 0
3 years ago
A particle having a speed of 0.87c has a momentum of 10-16 kg·m/s. what is its mass?
cupoosta [38]
The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
p=mv
In our problem, we know p=10^{-16}~Kg \cdot m/s and v=0.87c=0.87\cdot 3\cdot10^8~m/s=2.61\cdot 10^8~m/s, and using the relationship mentioned above, we can find the mass m of the particle:
m= \frac{p}{v} =3.8\cdot10^{-25}~Kg
8 0
3 years ago
The graphic organizer above shows that the properties of waves are influenced by the energy of waves. Name 2 properties of waves
Stells [14]
Amplitude: the height of the wave<span>, measured in meters
</span><span>Wavelength: the distance between adjacent crests, measured in meters
</span>
3 0
2 years ago
The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0
3 years ago
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