Answer:
See the answer below
Explanation:
The optimal conditions for high biodiversity seem to be a <u>warm temperature</u> and <u>wet climates</u>.
<em>The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 </em>
<em> and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots. </em>
Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 and wet environment with annual precipitation of not less than 262 cm.
The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.
Answer:
If she stands on the North side of a river flowing to the East at 5 mph,
she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have
x^2 + 5^2 = 10^2 where x is her speed directly across the river
x = (75)^1/2 = 8.66 mph towards the South
sin theta = 5 / 10 = 1/2
She must angle the boat at 30 deg from straight South
Answer:
the propagation velocity of the wave is 274.2 m/s
Explanation:
Given;
length of the string, L = 1.5 m
mass of the string, m = 0.002 kg
Tension of the string, T = 100 N
wavelength, λ = 1.5 m
The propagation velocity of the wave is calculated as;
Therefore, the propagation velocity of the wave is 274.2 m/s
Answer:
c) It has a greater frequency than red light but a smaller frequency than blue light.
Explanation:
According to the relation:
c = frequency × Wavelength
The higher the frequency, the lower the value of wavelength
The order of wavelength is:
Violet < Indigo < Blue < Green < Yellow < Orange < Red
Stated above, frequency is inversely proportional to the wavelength. Thus, the order of wavelength is:
Violet > Indigo > Blue > Green > Yellow > Orange > Red
Thus,
<u>Green light has lower frequency than blue light and higher than red light.</u>
Answer:
y_red / y_blue = 1.11
Explanation:
Let's use the constructor equation to find the image for each wavelength
1 /f = 1 /o + 1 /i
Where f is the focal length, or the distance to the object and i the distance to the image
Red light
1 / i = 1 / f - 1 / o
1 / i_red = 1 / f_red - 1 / o
1 / i_red = 1 / 19.57 - 1/30
1 / i_red = 1,776 10-2
i_red = 56.29 cm
Blue light
1 / i_blue = 1 / f_blue - 1 / o
1 / i_blue = 1 / 18.87 - 1/30
1 / i_blue = 1,966 10-2
i_blue = 50.863 cm
Now let's use the magnification ratio
m = y ’/ h = - i / o
y ’= - h i / o
Red Light
y_red ’= - 5 56.29 / 30
y_red ’= - 9.3816 cm
Light blue
y_blue ’= 5 50,863 / 30
y_blue ’= - 8.47716 cm
The ratio of the height of the two images is
y_red ’/ y_blue’ = 9.3816 / 8.47716
y_red / y_blue = 1,107
y_red / y_blue = 1.11
