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8090 [49]
3 years ago
15

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa

to 4.00×105 Pa. The second process is a compression to a volume of 5.00×10−2 m3 at a constant pressure of 4.00×105 Pa. Find the total work done by the gas during both processes.
Physics
1 answer:
Alborosie3 years ago
7 0
<h2>The work done = - 2 x 10⁴ J</h2>

Explanation:

In the first case , the volume is kept constant and pressure varies .

In isothermal process  , the work done

W₁ = V x ΔP

here V is the volume of gas and ΔP is the change in pressure

Thus W₁ = 0

Because there is no change in volume , therefore displacement is zero .

In second case pressure is constant , but volume changes

Thus W₂ = P x ΔV

here P is the pressure  and ΔV is the change in volume

Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

The total work done W = - 2 x 10⁴ J

Because the work done in compression is negative .

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II. IDENTIFY THE TYPE OF FORCE
uysha [10]

Answer:

Explanation:

1. Frictional force is responsible for running of car and buses on roads.

2.Gravitational force exists between astronauts in space.

3. Magnetic for is responsible to attract the iron objects using a magnet

4.Electrostatic force is responsible for fiber to stick on the skin.This force occurs due to the presence of charge.

5.When a person is pushing a trolley then object experience a normal reaction from ground.

6.Gravitational force makes the planet to move in their orbits.

4 0
3 years ago
A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

<u>Average acceleration</u>

F=ma, so a=\frac{F}{m_{b}}.

Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
A child lies on his back and raises his head up off the floor. When doing so, the total tension force in his neck muscles is 51.
Margarita [4]

Answer:

The total tension in the child's neck muscle, T = 56.51 N

Explanation:

Let m = mass of the child's neck

Radius of the curve, r = 2.40 m

The child's speed, v = 3.35 m/s

The tension force on the child's neck when he raises his head up off the floor, T_{f} = 51.0 N

The tension force on the child's neck when he raises his head from the wall of the slide, T_{s} = ?

T_{f} = mg\\g = 9.8 m/s^2\\51 = m * 9.8\\m = 51/9.8\\m = 5.2 kg

Since he makes a circular turn in water, the radial acceleration can be given by the equation:

a_{r} = \frac{v^{2} }{r} \\a_{r} = \frac{3.35^{2} }{2.4}\\a_{r} = 4.68 m/s^2

T_{s} = ma_{r} \\T_{s} = 5.2 * 4.68\\T_{s} = 24.336 N

The total tension in the child's neck muscle till be calculated as:

T = \sqrt{T_{f} ^{2} + T_{s} ^{2} } \\T = \sqrt{51 ^{2} + 24.336^{2} }\\T = \sqrt{2601 + 592.24 }\\T = 56.51 N

5 0
2 years ago
Cuál es la resistencia de un foco que está conectado a una fuente de 20 v por el cuál pasa una intensidad de 20 mA
LUCKY_DIMON [66]

Answer:

The resistance of the bulb is 1000 ohms.

Explanation:

What is the resistance of a bulb that is connected to a 20 v source through which a current of 20 mA passes

We have,

Voltage of the bulb is 20 V and the current passes is 20 mA.

It is required to find the resistance of the bulb. The relation between resistance, current and voltage is given by using Ohm's law as :

V =  IR

R is resistance

R=\dfrac{V}{I}\\\\R=\dfrac{20}{20\times 10^{-3}}\\\\R=1000\ \Omega

So, the resistance of the bulb is 1000 ohms.

7 0
2 years ago
When a certain metal is illuminated with light of frequency 3.0×1015 hz, a stopping potential of 7.27 v is required to stop the
slega [8]
Hf = Ф + Kmax

Where,
h = 4.14*10^-15 eV.s
f = 3.0*10^15 Hz
Kmax = 7.27 eV
Ф = ?

Therefore,
Ф= hf-Kmax = 4.14*10^-5*3.0*10^15 - 7.27 = 5.15 eV
7 0
2 years ago
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