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
First of all, don't forget that the sun is 400 times farther from us than the moon is. That fact alone tells us that anything on the earth is attracted to each kilogram of the moon with a force that's 160,000 times stronger than the force that attracts it to each kilogram of the Sun.
But more to your point ... The tides ARE greatly influenced by the sun. That's why tides are considerably higher at New Moon, when the sun and moon are both pulling in the same direction.
Answer:
b. calculate the constant force exerted on the pole vaulter due to the collision
mbgiclbxhkr and you can see the pole and you are not to be a great friend of the day and night sweets is my first day of your life and my friends is my friend
Answer:
b) √[(kx²/m) - 2gx]
Explanation:
The energy at the lowest point is equal to:
where:
Eelas = elastic energy [J]
k = spring constant [N/m]
x = extension of the spring [m]
We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.
