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11111nata11111 [884]
2 years ago
14

If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is th

e increase in the internal energy of the gas?
Physics
1 answer:
cluponka [151]2 years ago
7 0

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

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13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
2 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1) . The board rests on a frictionless horizontal s
chubhunter [2.5K]

Explanation:

Whole system will accelerate under the action of applied force. The box will experience the force against the friction and when this force exceeds then the box will move. so

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The applied force is given by

F = (m1 + m2)×a so

F = μs×g×(m1+m2)

3 0
2 years ago
A bicyclist is riding to the left with a velocity of 14 \,\dfrac{\text m}{\text s}14 s m ​ 14, start fraction, start text, m, en
Gnesinka [82]

Answer:

-2.0 m/s²

Explanation:Acceleration is the rate of change of velocity.

\begin{aligned}a&=\dfrac{\text{Change in velocity}}{\text{Change in time}}\\ \\ &=\dfrac{v_f-v_i}{\Delta t} \end{aligned}

a

​

 

=

Change in time

Change in velocity

​

=

Δt

v

f

​

−v

i

​

​

​

Hint #22 / 3

We can calculate the bicyclist's acceleration from the final velocity v_fv

f

​

v, start subscript, f, end subscript, initial velocity v_iv

i

​

v, start subscript, i, end subscript, and time interval \Delta tΔtdelta, t.

\begin{aligned}a&=\dfrac{v_f-v_i}{\Delta t}\\ \\ &=\dfrac{-21\,\dfrac{\text m}{\text s}-(-14\,\dfrac{\text m}{\text s})}{3.5\,\text s}\\ \\ &=-2.0\,\dfrac{\text m}{\text s^2}\end{aligned}

a

​

 

=

Δt

v

f

​

−v

i

​

​

=

3.5s

−21

s

m

​

−(−14

s

m

​

)

​

=−2.0

s

2

m

​

​

Hint #33 / 3

The acceleration of the bicyclist is -2.0\,\dfrac{\text m}{\text s^2}−2.0

s

2

m

​

minus, 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.

5 0
2 years ago
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

8 0
2 years ago
A wheel with rotational inertia 0.04 kg•m2 and radius 0.02 m is turning at the rate of 10 revolutions per second when a friction
Rus_ich [418]

Answer:

-78.96 J

Explanation:

The workdone by the torque in stopping the wheel = rotational kinetic energy change of wheel.

So W = 1/2I(ω₁² - ω₀²) where I = rotational inertia of wheel = 0.04 kgm², r = radius of wheel = 0.02 m, ω₀ = initial rotational speed = 10 rev/s × 2π = 62.83 rad/s, ω₁ = final rotational speed = 0 rad/s (since the wheel stops)

W = 1/2I(ω₁² - ω₀²) = 1/2 0.04 kgm² (0² - (62.83 rad/s)²) = -78.96 J  

7 0
2 years ago
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