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Y_Kistochka [10]
3 years ago
14

an ideal gas system undergoes an adiabatic process in which it expands and does 20 J of work on its environment. how much energy

is transferred into the system as heat
Physics
1 answer:
ehidna [41]3 years ago
8 0

Answer:

Zero.

Explanation:

An adiabatic process is one in which there is no exchange of heat energy. Therefore, in an adiabatic process, heat is neither added to the system not it is removed from the system.

The work done by the gas on the environment is 20 J. This energy is equal to the change in internal energy for an adiabatic process.

Therefore, for an ideal gas to undergo an adiabatic process in which it expands and does 20 J of work on its environment, the heat exchange is zero.

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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

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6 0
2 years ago
A spiral spring has a length of 14 cm when a force of 4 N is hung on it. A force of 6 N extends
stira [4]

Answer:

bussy

Explanation:

shart

5 0
2 years ago
A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi
vodka [1.7K]

Answer:

Explanation:

a ) Time period  T = 2 s

Angular velocity ω = 2π / T

=  2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ =  I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

w_2=\frac{356.25\times3.14}{256.25}

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² -  1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.

7 0
3 years ago
A 6 kg bowling ball is accelerated at a rate of 2.3 m/s² down the lane. How much force was necessary to produce this acceleratio
Maurinko [17]
The answer is D hoped this helped
4 0
3 years ago
A load of 800 N is lifted using a block and tackle having 5 pulleys. If the applied effort is 200 N, calculate
ser-zykov [4K]

Explanation:

Load=800N

Effort=200N

1. Mechanical Advantage = LOAD/EFFORT

= 800N/200N

= 4

2 Velocity Ratio = no. Of pulleys =5

3. Efficiency = Mechanical advantage / velocity ratio × 100%

= (4/5)×100%

=80%

4. output work= load×load distance

= 800N × 5m

= 4 × 1000J

5. Efficiency = (output work/input work) ×100%

Or, 80% = (4000J/input work) ×100%

Or, 80%/100% = 4000J/inputwork

Or, 4/5 = 4000J/inputwork

Or, input work =4000J × 5/4

Input work = 5×1000J

I hope it helped! ;-)

8 0
3 years ago
Read 2 more answers
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