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3 years ago

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A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

g released?

1 answer:

mars1129 [50]3 years ago

4 0

<span>(M G H)=(0.5 x 9.8 x 10) = 49 joules.</span>

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Something plus something equals total energy

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A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8

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Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet $m =6.75\times10^{24}\ kg$

Mass of star $m'=2.75\times10^{29}\ kg$

Radius of the orbit $r =8.05\times10^{7}\ km$

Using centripetal and gravitational force

The centripetal force is given by

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$F=m\omega^2r$

We know that,

$\omega=\dfrac{2\pi}{T}$

$F=m(\dfrac{2\pi}{T})^2r$ ....(I)

The gravitational force is given by

$F = \dfrac{mm'G}{r^2}$ ....(II)

From equation (I) and (II)

$m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}$

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

$T^2=\dfrac{4\pi^2R^3}{m'G}$

$T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}$

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3 years ago

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The total oxygen atoms that are exprected in the vinegar and baking soda reaction is three oxygen atoms.

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3 years ago

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What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen a

Fiesta28 [93]

Answer:

a) $E_photon =0.306 eV$

b) $E_photon =0.166 eV$

Explanation:

The energy of the photon (E) for $n^th$ orbit of the hydrogen atom is given as:

$E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$

where,

$E_o$ = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

$E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})$

$E_photon =0.306 eV$

b) Now for the transition from n = 5 to n = 6

$E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})$

$E_photon =0.166 eV$

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3 years ago