Answer:
The value is 
Explanation:
From the question we are told that
The distance of separation is 
The current on the one wire is 
The current on the second wire is 
Generally the magnitude of the field exerted between the current carrying wire is

Here
is the magnetic field due to the first wire which is mathematically represented as

Here
is the distance to the half way point of the separation and the value is

is the magnetic field due to the first wire which is mathematically represented as

Here
is the distance to the half way point of the separation and the value is
This means that 
So

=> 
=> 
=> 
Answer: 8000N
Explanation: since it is frictionless that means it has 100% efficiency therefore the mechanical advantage is 1 meaning the load equals to the effort
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
Answer:
(C) The frequency decrease and intensity decrease
Explanation:
The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.
if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.
The frequency is the inverse from the wavelength, so the frequency heard will increase.
The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.