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Y_Kistochka [10]
3 years ago
14

What is the purpose of a free body diagram?

Physics
1 answer:
leonid [27]3 years ago
8 0

Answer:

i think the answer is c

Explanation:

zrrxtcyvjvugugyctcyvjv

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The ____ of a position-time graph represents an object’s velocity.
Marianna [84]

Answer:

The slope of a position-time graph represents an object’s velocity.

Explanation:

In a position-time graph, the values on the x-axis represent the time, while the values on the y-axis represent the position of the object.

Velocity is defined as the ratio between the displacement of an object and the time taken:

v=\frac{\Delta s}{\Delta t}

However, we can see that this definition corresponds to the slope of the curve in a position-time graph. In fact:

\Delta s, the displacement, corresponds to the difference in position, so the difference between the values on the y-axis: \Delta s=y_2 -y_1

\Delta t, the time interval, corresponds to the difference in times, so the difference between the values on the x-axis: \Delta t= t_2 -t_1=x_2 -x_1

So, the velocity is

v=\frac{\Delta s}{\Delta t}=\frac{y_2 -y_1}{x_2 -x_1}

which corresponds to the slope of the curve.

3 0
3 years ago
Read 2 more answers
The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be
Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
  • It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
  • Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
  • The time period of geosynchronous orbit is 24 hours or 1440 minutes.
  • As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
  • If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
  • a1= (T1/T2)⅔×a2

           = (1440/90)⅔×6780

           = 43,090 km

  • Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

#SPJ4

   

6 0
1 year ago
In general, which statement does your data support ?
hjlf

Answer:

A

Explanation:

As distance increases, velocity increases.

5 0
2 years ago
What is the difference between amplitude and wavelength?
jasenka [17]

Answer:

Letter b is wavelength. Letter a is amplitude.

Explanation:

Let's imagine a simple experiment. Imagine you have a long thick rope which one end is at your hands, and you start an oscillatory motion in it, moving your hand up and down. Then a friend of you take a picture of the rope in motion, looking at the rope laterally. Now let's find the wavelength and amplitude. Amplitude is "The distance from the center of the oscillation of the rope (when the rope was not in motion) to its high or low point", or the vertical displacement, in our experiment. On the other hand, wavelength is "The distance between one high point /low point and the next high point /low point". Take a look at a photo of a wave in your textbook and you will find the answer as well. ; )

3 0
3 years ago
The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o
Anuta_ua [19.1K]

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

E_n=\frac{-13.6eV}{n^2}

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

E_{k}=11.1eV-10.2eV=0.9eV

6 0
3 years ago
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