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romanna [79]
3 years ago
15

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?
Physics
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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your answer should be astronomy

Explanation:

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Something plus something equals total energy
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potential plus kinetic

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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
Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planetm =6.75\times10^{24}\ kg

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

Radius of the orbitr =8.05\times10^{7}\ km

Using centripetal and gravitational force

The centripetal force is given by

F = \dfrac{mv^2}{r}

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}}

T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}

T =3.34\times10^{7}\ s

T= 387.62\days

Hence, The orbital period of the planet is 387.62 days.

4 0
3 years ago
Read 2 more answers
Vinegar (C2H4O2) and baking soda (NaHCO3) react to form sodium acetate, water, and carbon dioxide. C2H4O2 + NaHCO3 ? How many to
aliya0001 [1]

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


Answer: B. 3




why;live

5 0
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

7 0
3 years ago
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