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s344n2d4d5 [400]
2 years ago
8

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

and energy increases. increases and energy decreases
Physics
2 answers:
lora16 [44]2 years ago
4 0

Answer:

I TOOK THE TEST! the answer is INCREASES AND ENERGY INCREASES!

frozen [14]2 years ago
3 0
Bohr's equation for the change in energy is
\Delta E= \frac{hc}{\lambda}
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
     TRUE

b. As λ increases, f increases and ΔE increases.
    FALSE

c. As λ increases, f increases and ΔE decreases.
    FALSE

Answer: 
As the wavelength increases, the frequency decreases and energy decreases.

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Photosynthesis is the process plants use to synthesize food from carbon dioxide and water. Sunlight is used as a energy source. Photosynthesis releases oxygen as a byproduct.

At low temperatures, between 0 and 10 degrees Celsius – the enzymes that carry out photosynthesis do not work efficiently, and this decreases the photosynthetic rate.

At medium temperature above 10 degrees Celsius to below 40 degrees Celsius (e.g 27 degrees Celsius), the photosynthetic enzymes work at their optimum levels, so photosynthesis proceeds.

At a temperature above 40 degrees Celsius, the enzymes that carry out photosynthesis lose their shape and functionality, and the photosynthetic rate declines rapidly.

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2 years ago
What's the momentum of a 3.5-kg boulder rolling down hill at 5 m/s
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Identify the method of thermal energy transfer at work in hot air balloons. Explain how thermal energy is transferred in this sc
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A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
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24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

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3 0
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