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elena-14-01-66 [18.8K]
3 years ago
14

As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air

acts as an ideal gas, calculate the work done per mol of the air.
A. -80.35 J
B. 80.35 J
C. -811.93 J
D. 811.93 J
(please show calculation)
can use this formula W=nRT ln(p1/p2)​
Physics
1 answer:
MArishka [77]3 years ago
7 0

This question involves the concepts of work done, pressure, and temperature.

The work done per mol of the air is "-724.71 J/mol".

Using the given formula for work done:

W=RTln\frac{P_1}{P_2}

where,

W = work done per mol = ?

R = universal gas constant = 8.314 J/mol.k

T = absolute temperature = 30°C + 273 = 303 k

P₁ = initial pressure = 30 KPa

P₂ = final pressure = 40 KPa

Therefore,

W= (8.314\ J/mol.k)(303\ k)ln\frac{30\ KPa}{40\ KPa}\\

<u>W = -724.71 J/mol</u>

<u></u>

Learn more about pressure here:

brainly.com/question/23358029?referrer=searchResults

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3 years ago
Pam has a mass of 42.4 kg and she is at rest on smooth, level, frictionless ice. Pam straps on a rocket pack. The rocket supplie
fredd [130]

Answer:

F = 2349.6 N

Explanation:

We can solve this exercise using the relationship of momentum and momentum

         I = Δp

         I = F t

As the woman accelerates at a distance of 29.1 m to go from rest to 56.8 m / s, we can use the kinematics to find the acceleration

       v² = v₀² + 2 a x

       v₀ = 0

       a = v / 2x

       a = 56.8 2/2 29.1

       a = 55.43 m / s²

Let's look for the time you need to get this speed

      v = v₀ + a t

      t = v / a

      t = 56.8 / 55.43

      t = 1,025 s

Let's clear the average force momentum from the momentum

      F t = m v- m v₀

      F = mv / t

      F = 42.4 56.8 / 1.025

      F = 2349.6 N

3 0
3 years ago
A projectile is fired straight upward at 173 m/s. how fast is it moving at the instant it reaches the top of its trajectory?
Fittoniya [83]
The top of the trajectory is the point where it changes from rising to falling. At that exact instant, its vertical speed is zero.
8 0
3 years ago
Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N
LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

x(t) = A\cos(\omega t + \phi)

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude.  According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1.  We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3

Then the velocity at that point is

v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)

The maximum speed is where the acceleration is equal to zero:

0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

-\omega A(0.866) \neq -\omega A

2. The maximum acceleration is at the amplitude.

A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A

And the acceleration at A/2 is

\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0
3 years ago
HELP HELP..................... ​
tatiyna

Hi,

Your question is "which graph represents uniform motion", the answer would be option 1

Have a good one!

6 0
4 years ago
Read 2 more answers
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