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3 years ago

_____ are pictures of relationships

Physics

2 answers:

Alik [6]3 years ago

6 0

Answer:

No idea

Explanation:

butalik [34]3 years ago

4 0

The blank in the question can be filled with the word, “Graph”. Therefore, Graphs are the pictures which are in relationships.

<u>Explanation: </u>

Graph usually represents a set of data which is nonlinear in occurrence and has some relationship between the two given data. And as graph are pictorial representation, it is simply assumed as the pictures of relationships.

For example, a graph can be drawn for the set of data for the presence of number of students of all the sections of the particular class of a school, as they are relative. But making the graph for number of students in all section of all class but different school cannot be done as non-relative.

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Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

2 years ago

an electron, a proton and a deuteron move in a magnetic field with same momentum perpendicularly. the ratio of the radii of thei

wel

If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:

1: √2 : 1

<h3>How is the ratio of the perpendicular parts obtained?</h3>

To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle = 4m.

The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:

r = √m/e : √2m/e : √4m/2e

When resolved, the resulting ratios will be:

1: √2 : 1

Learn more about the radii of their circular paths here:

brainly.com/question/16816166

#SPJ4

6 0

1 year ago

If the center of mass passes outside the area of support of an object, what will happen to it?

defon

Answer:

If a vertical line extending down from an object's CG extends outside its area of support, the object will topple

Explanation:

We can understand better this situation using a diagram with the forces acting on it.

In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.