Answer:
Explanation:
Given
Temperature of Room 
Area of Person 
Temperature of skin 
Heat transfer coefficient 
Emissivity of the skin and clothes 

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection
Heat transfer due radiation 
where 


Heat Transfer due to convection is given by




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

=> 
=> 
=> 
Factors that affect heat transfer are:
1) Difference in temperature,
2) Mass of the object
3) Specific heat of the object
Hope this helps!
Explanation:
Work = force × displacement
532 J = 48 N × d
d ≈ 11 m