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3 years ago

9

How do scientists construct models of atoms and molecules?

1 answer:

9966 [12]3 years ago

6 0

Mr. Roentgen's x-rays allowed scientists to measure the size of the atom. The x-rays were small enough to discern the atomic clouds. This was done by scattering x-rays from atoms and measuring their size just as Rutherford had done earlier by hitting atoms with other nuclei starting with alpha particles.

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A wave with a greater amplitude will transfer . . . . \

BabaBlast [244]

less mass is more mass but less energy in more mass. less mass has more energy

6 0

3 years ago

An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i

Svetllana [295]

Answer:

$v \approx 9.312\,\frac{m}{s}$

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

$K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}$

$K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot s\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cdot s \cos \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)$

$v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}$

$v \approx 9.312\,\frac{m}{s}$

3 0

3 years ago

Question 7 (1 point)

Temka [501]

Answer:

Forms over water, warm humid air mass, it's a polar air mass

Explanation: I think that's right sorry if it's not..

GL! :)

4 0

2 years ago

Which activity is the best example of cardiovascular and strength training exercises work together

Nataly [62]

it got to be jumping jacks that all i got

4 0

3 years ago

Read 2 more answers

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

or

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago