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Nata [24]
3 years ago
13

A photographic film can be exposed (affected) by light with a minimum energy of 2.65 x 10-19 J. How many of the colors of the vi

sible light can expose this film?
Physics
1 answer:
beks73 [17]3 years ago
5 0

Answer:

All of the colors of the visible light spectrum can expose the film

Explanation:

energy can be calculated with the following formula;

E=\frac{hc}{λ}  

where h is planck constant = 6.62 * 10^{-34}  

c is speed of light = 3* 10^{8}  

λ is wavelength  

visible light comprises of the ROYGBIV colors, which are red, orange yellow, green, blue, indigo and violet. For any of these colors to affect the film their energy must be greater than the minimum energy required to expose the film. However, violet with the smallest wavelength has the highest energy. we would calculate the energy of red color in the visible spectrum to see if it surpasses this minimum energy.

highest possible wavelength(red color) = 750*10^{-9}m  

E(red light)=\frac{6.62*10^{-34} *3*10^{8} }{750 * 10^{-9} }

E= 2.646 * 10^{-19} J

≈ 2.65 * 10^{-19} J

this energy is for the highest wavelength exhibited by red color. for lower wavelengths of red color in the spectrum the film will be exposed and also this indicates that other colors in the spectrum will excite the film as they have very much lower wavelengths

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Given what we know, the statement in this question can be considered as true, since Syncopation does in fact shift the beat off of regular rhythms.

<h3>What is Syncopation?</h3>
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For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

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