Answer:
I don't know sorry For this question
The frequency of the wave is 6800 Hz
<u>Explanation:</u>
Given:
Wave number, n = 20
Speed of light, v = 340 m/s
Frequency, f = ?
we know:
wave number = 

Therefore, the frequency of the wave is 6800 Hz
Answer:
270 m
Explanation:
Given:
v₀ = 63 m/s
a = 2.8 m/s²
t = 4.0 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (63 m/s) (4.0 s) + ½ (2.8 m/s²) (4.0 s)²
Δx = 274.4 m
Rounded to two significant figures, the displacement is 270 meters.
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
Thank you for reading.