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DerKrebs [107]
3 years ago
13

26. The equation E=hf describes the energy of each photon in a beam of light. If Planck’s constant, h, were larger, would photon

s of light of the same frequency be more energetic or less energetic. Be sure to explain using at least 3 content related sentences.
Physics
1 answer:
Elenna [48]3 years ago
6 0
The equation of the energy of a photon is E=h*f.

If we increase the Planck's constant h, the energy would increase. 

For example, lets double the value of Planck's constant and name it H:

H=2*h. Now lets put that into the equation for energy that we will call E₂:

E₂=H*f=2*h*f=2*E. 

So we can clearly see that E₂=2*E or that if we double Planck's constant, the energy also doubles. 
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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

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r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

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8 0
3 years ago
When a person speaks, a sound intensity is generated that is 600 times greater than when the person whispers. What is the differ
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Answer:

Originally :  Level = log I / I0

Currently: Level = 10 log I / I0

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Note the term 1 bel = 10 decibels

5 0
2 years ago
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Answer:

K.E =  \frac{1}{2} m {v}^{2}  \\  {v}^{2}_i  =  {v}^{2} _x + {v}^{2} _y \\  =  {(6)}^{2}  +  {( - 2)}^{2}  = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (40) = 60J \\  \\ {v}^{2}_f  =  {v}^{2} _x + {v}^{2} _y \\  =  {(8)}^{2}  +  {(4)}^{2}  = 80m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\  = 120J - 60J \\  = 60J

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The electric current running through the wire coil in an electric motor exerts force directly onto
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C) A powerful magnet
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