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miss Akunina [59]

3 years ago

8

What is the energy of a photon with a frequency of 5 x 10^14 Hz?

1 answer:

zloy xaker [14]3 years ago

8 0

Answer:

$3.315 x 10^-19 J$

Explanation:

Using Eisten's formula, energy is a product of Plank's constant and photon frequency

E=hf

Where f represent frequency, h is Plank's constant and E is energy.

Substituting frequency with 5 x 10^14 Hz and

Planck's constant is 6.63 x 10^-34 J•s.

$E=6.63 x 10^{-34} *5 x 10^{14}=3.315x10^{-19} J$

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${ \bf{ \underbrace{Answer :}}}$

$\sf\red{B. \:They're\: inert.}$ ✅

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2 years ago

The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.

kiruha [24]

Answer:

$3.1\cdot10^{23}\:\mathrm{kg}$

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

$g_P=G\frac{m}{r^2}$ ., where $g_P$ is acceleration due to gravity at the planet's surface, $G$ is gravitational constant $6.67\cdot 10^{-11}$ , $m$ is the mass of the planet, and $r$ is the radius of the planet.

Since acceleration due to gravity is given as $m/s^2$ , our radius should be meters. Therefore, convert $2400$ kilometers to meters:

$2400\:\mathrm{km}=2,400,000\:\mathrm{m}$ .

Now plugging in our values, we get:

$3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2}$ ,

Solving for $m$ :

$m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}$ .

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2 years ago