Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
10 minutes are the same as 600 seconds.
If you run 2 meters in 1 second then you run 2 * 600 meters in 600 seconds.
Answer:
a. Light pollution refers to light used for human activities that brightens the sky and hinders astronomical observations.
Explanation:
Light pollution is due to the excessive and misdirected use of artificial light. Light bulbs are often design in an incorrect way, since a great part of its light is not completely directed to the ground and an important percentage is emitted to the sky in where will be scattered and reflected back to ground by the particles in the atmosphere. That brings as an effect a sky glow, therefore the visibility of astronomical objects will be extremely reduce.
Hence, professional astronomical research and amateur observations will be affected. Light pollution has a negative impact on bird migration at night and in the health of difference species, humans also.
Answer:
I = 5[amp]
Explanation:
Electrical power is defined as the product of voltage by current.
where:
P = power = 1150 [W]
V = voltage = 230 [V]
I = current [amp]
Now replacing:
![1150=230*I\\I=1150/230\\I=5[amp]](https://tex.z-dn.net/?f=1150%3D230%2AI%5C%5CI%3D1150%2F230%5C%5CI%3D5%5Bamp%5D)
A 15 [amp] fuse must be used. Always the fuse must be larger than the operating current, to protect the equipment from very high currents. above 15 [amp]
Something that goes round in circles could have a high speed but zero average velocity.
