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1 year ago

Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.

Physics

1 answer:

mote1985 [20]1 year ago

7 0

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The expression is $W_c = P_o V_o ln (R_v)$

Explanation:

Generally smallest workdone done by a gas is mathematically represented as

$dW = PdV$

Generally for an isothermal process

$PV = nRT = constant$

=> $P = \frac{nRT}{V}$

Generally the total workdone is mathematically represented as

$W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV$

=> $W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV$

=> $nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.$

=> $W_c = nRT [ln(V_f) - ln(V_o)]$

=> $W_c = nRT ln \frac{V_f}{V_o}$

From the question $\frac{V_f}{V_o } = R_v$

=> $W_c = P Vln (R_v)$

at initial state

$W_c = P_o V_o ln (R_v)$

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A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the

Blizzard [7]

<h3><u>Answer;</u></h3>

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant

F = k*y

Therefore; k = F/y = m*g/y

= 0.40kg*9.8m/s^2/(0.76 - 0.41)

= 11.2 N/m

Energy is conserved

Let A be the maximum displacement

Therefore; 1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2

Thus; A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

= sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

= 0.846 m

Thus; the length will be 0.41 + 0.846 = 1.256 m

6 0

1 year ago

WILL GIVE BRAINLIEST AND 100 POINTS! NEED THIS ASAP!

lorasvet [3.4K]

Answer:

6.57, 1.64, .88

Explanation:

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8 0

1 year ago

Physics help, please? :)

Svetllana [295]

Answer:

Explanation:

3. Newton’s third law explains how every action has an equal but opposite reaction, meaning that forces comes in pairs. While the locomotive’s wheels are pushing back against the ground as the action force, the ground is producing a reaction force towards the locomotive, propelling it forward. Another pair of forces that act on the locomotive is gravity and normal force. While gravity is pulling the locomotive towards the ground, the normal force the ground exerts on the locomotive is why the locomotive doesn’t fall through the ground.

4. The force of Earth’s gravity on the Sun is weaker than the force of the Sun’s gravity on Earth. The Sun’s attraction affects the motion of Earth more than the Earth’s attraction affects the Sun’s motion because according to Newton’s second law, force has mass as one of its factors. The Sun has a significantly higher mass than Earth, meaning that its force of gravity would also be significantly higher. Newton’s third law is why the Earth doesn’t get marginally closer to the Sun, stating that every action has an equal and opposite reaction. As the Sun is pulling Earth towards itself, Earth is pulling away from the Sun.

8 0

1 year ago

Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

<em>x= 0.67 cm to the right of q1</em>

Explanation:

<u>Electrostatic Forces</u>

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$

Assuming the positive sign :

$\displaystyle 0.02-x= 2x$

$\displaystyle 3x=0.02$

$\displaystyle x=0.00667\ m$

$x=0.67\ cm$

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

$\displaystyle 0.02-x=-2x$

$\displaystyle x=-0.02\ m$

$\displaystyle x=-2\ cm$

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

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1 year ago

An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and

torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0

1 year ago