Answer:
206.8965517 n
Explanation:
First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.
Given: A cubic tank holds 1,000.0 kg of water.
Mass of water in tank (m) = 1000.0 kg
Density of water (d) = 1000.0 kg /m³
Concept: Volume(V) = Mass / Density
Since the tank holds these water in it so the volume of water will be equal to the volume of the tank.
Hence, the volume of the tank = Mass of water / Density of water
or, = 1000.0 kg / 1000.0 kg m⁻³
or, = 1.0 m³
Since tank is cubical in shape. Let its side be 'x'
The volume of tank (x³) = 1.0 m³
or. side of tank (x) = 1.0 m
Hence, the dimensions of the tank will be 1.0 m.
C im pretty sure?.........
Answer:
6.71×10⁻⁷ m
Explanation:
Using thin film constructive interference formula as:
<u>2×n×t = m×λ</u>
Where,
n is the refractive index of the refracted surface
t is the thickness of the surface
λ is the wavelength
If m =1
Then,
2×n×t = λ
Given that refractive index pf the oil is 1.22
Thickness of the oil = 275 nm
Also, 1 nm = 10⁻⁹ m
Thickness = 275×10⁻⁹ m
So,
Wavelength is :
<u>λ= 2×n×t = 2× 1.22 × 275×10⁻⁹ m = 6.71×10⁻⁷ m</u>
Answer:
a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m
b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m
Explanation:
(a) Radiowave wavelength= λ = c/f
As we know, Radiowave speed in the air = c = 3 x 10^8 m/s
f = frequency = 604 kHz = 604 x 10^3 Hz
Hence, wavelength = (3x10^8/604x10^3) m
λ
= 496.69 m
So the height of the antenna BROADCASTING AT 604 kHz = λ /4 = (496.69/4) m
= 124.17 m
(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz (1kHZ = 1000 Hz)
Hence, wavelength = λ = (3 x 10^8/1710 x 10^3) m
λ= 175.44 m
So, height of the antenna = λ /4 = (175.44/4) m
= 43.86 m