Answer:
D. Energy is converted between kinetic and gravitational potential energy.
Explanation:
(a) 
The relationship between frequency and wavelength of an electromagnetic wave is given by

where
is the speed of light
is the frequency
is the wavelength
In this problem, we are considering light with wavelength of

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

(b) 
The period of a wave is equal to the reciprocal of the frequency:

And using
as we found in the previous part, we can find the period of this wave:

Answer:
0.000000002 m=2.0*10⁻⁹ m
Explanation:
Scientific notation allows us to write very large or very small numbers in abbreviated form. This notation simply consists of multiplying by a power of base 10 with a positive or negative exponent.
A number written in scientific notation has the form:
a*10ⁿ
where:
- the coefficient a has a value such that 1 ≤ a <10
- n is an integer. Represents the number of times the decimal point is shifted. It is always a whole number, positive if it is shifted to the left, negative if it is shifted to the right.
So to write the number 0.000000002 in scientific notation, the following steps are performed:
- The decimal point is moved to the right as many spaces until it reaches the right of the first digit.
- This number is then written, which will be the coefficient a in the expression of the previous product. So a=2.0
- The base 10 is written with the exponent equal to the number of spaces that the comma moves. So n=9. But this is a negative number because the comma shifts to the right.
So, you get: <u><em>0.000000002 m=2.0*10⁻⁹ m</em></u>
Answer:
= 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
we substitute
5/9 = 