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Afina-wow [57]
3 years ago
15

A rigid tank contains 1.80 moles of helium, which can be treated as an ideal gas, at a pressure of 27.3 atm. While the tank and

gas maintain a constant volume and temperature, a number of moles are removed from the tank, reducing the pressure to 5.20 atm. How many moles are removed?
Physics
1 answer:
maria [59]3 years ago
8 0

Answer:

Has been removed 1.458 moles.

Explanation:

n1= 1.8 mol

p1= 27.3 atm

p2= 5.2 atm

n2= ?

n2= n1 * p2/p1

n2= 0.342 moles

Δn= n1-n2

Δn= 1.458 moles

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The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.
Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

  • Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

<u>Conversion:</u>

1 day = 24 hours

24 hours to minutes = 60 × 24 = 1440 minutes

1440 + 37 = 1477 \; minutes

1 minute = 60 seconds

1477 minute = X seconds

Cross-multiplying, we have:

X = 60 × 1477

X = 88620 seconds

Now, we can find the frequency of rotation of Mars by using the formula:

Frequency = \frac{1}{Period}\\\\Frequency = \frac{1}{88620}

<em>Frequency </em><em>of rotation</em> = <em>0.0000113 Hertz</em>

Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

Read more: brainly.com/question/14708169

8 0
3 years ago
A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th
Usimov [2.4K]

Answer:

\omega_f = 3.584\ rad/s

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed  = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

\theta = \omega_i t + \dfrac{1}{2}\alpha t^2

5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2

\alpha= \dfrac{10}{2.8^2}

       α = 1.28 rad/s²

now, calculation of angular velocity

\omega_f = \omega_i + \alpha t

\omega_f =0 +1.28\times 2.8

\omega_f = 3.584\ rad/s

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0
3 years ago
A rock with a mass of 0.3 kg falls from the top of a cliff. If it takes the rock 2.5 s to reach the ground, what was the impulse
kaheart [24]
Impulse=force x time
force=mass x acceleration due to gravity
force=
300 \times 10 = 3000
impulse =3000 x 2.5= ( sorry i don't have a calculator right now so you must calculate this yourself)
I converted from kg to g because it is the standard.
Hope this helps you.
4 0
3 years ago
Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

7 0
1 year ago
A 300 g bird is flying along at 6.0 m/s and sees a 10 g insect heading straight towards it with a speed of 30 m/s. The bird open
Gelneren [198K]

Answer:

(a): The bird speed after swallowing the insect is V= 4.83 m/s

(b): The impulse on the bird is I= 0.3 kg m/s

(c): The force between the bird and the insect is F= 20 N

Explanation:

ma= 0.3 kg

va= 6 m/s

mb= 0.01kg

vb= 30 m/s

(ma*va - mb*vb) / (ma+mb) = V

V= 4.83 m/s (a)

I= mb * vb

I= 0.3 kg m/s  (b)

F*t= I

F= I/t

F= 20 N (c)

5 0
3 years ago
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