Answer:
Explanation:
Given
saturated air temperature by 
Dew point temperature is given by 
Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.
air continues to rise for 1400 m
i.e. change in temperature would be 
Final temperature 


Answer:
Given the area A of a flat surface and the magnetic flux through the surface
it is possible to calculate the magnitude
.
Explanation:
The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux
is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (
). So 1 Wb=1 T.m².
For a flat surface S of area A in a uniform magnetic field B, with
being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

We are told the values of
and B, then we can calculate the magnitude

Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The attached file has a detailed solution of the given problem.
Answer:
The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>
Explanation:
Given:
Upward direction is positive. So, downward direction is negative.
Tota time the ball remains in air (t) = 8.0 s
Net displacement of the ball (S) = Final position - Initial position = 0 m
Acceleration of the ball is due to gravity. So,
(Acting down)
Now, let the initial velocity be 'u' m/s.
From Newton's equation of motion, we have:

Plug in the given values and solve for 'u'. This gives,

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.