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3 years ago

IF THERE ARE ONLY 118 ELEMENTS, HOW DO YOU ACCOUNT FOR THE MANY MILLIONS OF THINGS THAT WE HAVE IN OUR UNIVERSE?

Physics

1 answer:

bogdanovich [222]3 years ago

3 0

Answer:

theres only 118 elements that are discovered. now that they're the only ones out there

Explanation:

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A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

Paul [167]

Answer:

The height of water in the column = 348.14 cm

Explanation:

Pressure:This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of pressure is N/m²

p = Dgh............... Equation 1

Where p = pressure, D = density, g = acceleration due to gravity, h = height.

From the question, the same pressure will support the column of mercury and water.

p₁ = p₂

Where p₁ = pressure of mercury, p₂ = pressure of water

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

h₂ = 348.14 cm

The height of water in the column = 348.14 cm

6 0

3 years ago

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point

seropon [69]

since centripetal acceleration is always towards the center of the circle

so at the given position where speed and acceleration is given the center coordinate will be towards the center of circle

also we know that

$a_c = \frac{v^2}{r}$

$11.6 = \frac{2.5^2}{R}$

$R = 0.54 m$

so the coordinates of the center will be

$(x, y) = (3.20 , (3.5 + 0.54))$

$(x, y) = (3.20, 4.04)$

so the coordinate is (3.20 m, 4.04 m)

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3 years ago

Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that

Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

$g_e$ = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

$m=M-0.1M\\\Rightarrow m=0.9M$

$g_e=G\frac{M}{r^2}$

$g=G\frac{0.9M}{r^2}$

Dividing the equations we get

$\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2$

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0

3 years ago

In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point

jarptica [38.1K]

The difference in electric potential energy between the two points is
$\Delta U = q \Delta V$
where q is the magnitude of the charge and $\Delta V$ is the electric potential difference.

But for energy conservation, the difference in electric potential energy $\Delta U$ between the two points is equal to the work done to move the charge between A and B:
$W=\Delta U$
so we have
$W=q \Delta V$

and by substituting the numbers of the problem, we find the value of $\Delta V$ :
$\Delta V = \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V$

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3 years ago

A plot of land has been surveyed for a new housing development with borders AB, BC, DC, and DA. The plot of land is a right trap

balandron [24]

Angle ABC is 119

Length of DC is 105

6 0

3 years ago

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