In that formula for Energy, 'F' is the frequency of the photon.
But <u>Frequency = (speed)/(wavelength)</u>, so we can write the
Energy formula as
E = h c / (wavelength) .
So the energy, in joules, of a photon with that wavelength, is . . .
E = (6.6 x 10⁻³⁴) x (3 x10⁸) / (that wavelength)
= <em>(1.989 x 10⁻²⁵) / (that wavelength, in meters) .</em>
Answer:
Explanation:
This is going to sound like an absurd answer, but sometimes physics can be a little strange.
This answer is weird because of the definition of displacement. It means the distance from the starting point to the ending point, disregarding what happened in between. The point is that the astronaut is at the starting point of his orbit. By definition the starting and ending points are the same. His displacement is 0.
So the answer is you have the greater displacement when you walked one way to school. The starting point and the ending point are different. You have gone further.
However just to make things a little nasty, when you walk home again, your displacement will be the same as the astronaut's -- 0 meters because you will be right back where you started from.
Answer:
Obviously Lengthen...
or 
Explanation:
As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:

we've the density of the water 


b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:



<u>Now the difference in the column is :</u>



Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) =
- K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ =
–K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J