Does anyone know this or done it? It’s for life skills and about MyPlate:)

As the collision frequency of gas particles increases, the mean free path of the gas particles ____

Mamont248 [21]

As the collision frequency of gas particles increases, the mean free path of the gas particles decreases.

<h3>Frequency </h3>

The number of times a repeated event occurs in a given amount of time is known as its frequency. It is also sometimes called "temporal frequency" to stress the contrast to "spatial frequency" and "ordinary frequency" to underline the contrast to "angular frequency." Hertz (Hz), which is equal to one (event) per second, are the units used to express frequency. The reciprocal of frequency, the period is the length of time occupied by one cycle in a repeating event. When describing the temporal rate of change seen in oscillatory and periodic phenomena like mechanical vibrations, audio signals (sound), radio waves, and light, frequency is a crucial parameter utilized in science and engineering.

Learn more about frequency here:

brainly.com/question/5102661

#SPJ4

4 0

1 year ago

The International Space Station (ISS) orbits Earth at an altitude of 400 km. Using this information, plus the mass and radius of

Thepotemich [5.8K]

Answer:

v = 7671.57 m/s

T = 1.55 hours

Explanation:

mass of Earth, M = 6 x 10^24 kg

Radius of earth, R = 6400 km = 6.4 x 10^6 m

height, h = 400 km

Velocity is given by

$v=\sqrt{\frac{GM}{R+h}}$

where, G be the universal gravitational constant.

G = 6.657 x 10^-11 Nm^2/kg^2

$v=\sqrt{\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{6800\times 10^{3}}}$

v = 7671.57 m/s

Let T b the period

$T=\frac{2\pi (R+h)}{v}$

$T=\frac{2\times 3.14(6800\times 1000)}{7671.57}$

T = 5566.53 second

T = 1.55 hours

5 0

3 years ago

A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati

Ainat [17]

Answer:

The asteroid's acceleration at this point is $2.71\ m/s^2$

Explanation:

The equation that governs the trajectory of asteroid is given by :

$x=6.5t-2.3t^3$

The velocity of asteroid is given by :

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2$

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

$6.5-6.9t^2=0\\\\t=0.971\ s$

Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t$

Put t = 0.971 s

$a=-13.8\times 0.197\\\\a=-2.71\ m/s^2$

So, the asteroid's acceleration at this point is $2.71\ m/s^2$ and it is decelerating.

6 0

3 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0

2 years ago

A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the

Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

$m_1v = m_1v_1 + m_2v_2$

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute $m_2 = m_1/2 \& v_1 = v/3$