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3 years ago

Give a few examples/ applications of the universal law of gravitation

1 answer:

8 0

Answer: 1) It keeps us on the earth so that we can live on the earth and not flying someone else in the atmosphere and space. 2) It maintains the motion of motion of all the planets around the sun and moon around the earth. 3) It pulls all the object towards the earth. 4) The flowing of water in the rivers and seas.

Explanation:

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Acrostic poem for cell theory. Especially theory

Sonja [21]

Answer:

In biology, cell theory is the historic scientific theory, now universally accepted, that living organisms are made up of cells, that they are the basic structural/organizational unit of all organisms, and that all cells come from pre-existing cells. Cells are the basic unit of structure in all organisms and also the basic unit of reproduction.

Explanation:

The three tenets to the cell theory are as described below:

All living organisms are composed of one or more cells.

The cell is the basic unit of structure and organization in organisms.

Cells arise from pre-existing cells.

There is no universally accepted definition of life. Some biologists consider non-cellular entities such as viruses living organisms,[1] and thus reasonably disagree with the first tenet. Throughout this article, it will lead you through the history of cell theory, how the discovery of cells was made possible, what the cell theory has become today and background information and history regarding other opposing concepts of cell theory.

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3 years ago

An ideal parallel - plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is conne

Svetradugi [14.3K]

Answer:

The capacitance is cut in half.

Explanation:

The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:

C = (\epsilon)*(A/d)

Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:

C_new = (\epsilon)*(A/2d)

C_new = (1/2)*[(\epsilon)*(A/d)]

Since C = (\epsilon)*(A/d)] we have:

C_new = (1/2)*C

4 0

3 years ago

You are looking at a yellow flower growing outside in the sunshine. Why does it look yellow?

Alchen [17]

Answer:

It looks yellow because that is the only (major) color reflected.

Visible spectra is from about 4000-7000 Angstroms (10^-10 m).

Red are longer wavelengths and blue are the shorter wavelengths.

The Sodium doublet (yellow) occurs around 5900 Angstroms.

3 0

2 years ago

Which unit is used to measure force

Ymorist [56]

Newtons (N) measure force

3 0

3 years ago

Read 2 more answers

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

4 years ago

Give a few examples/ applications of the universal law of gravitation​

Give a few examples/ applications of the universal law of gravitation