The international space station (ISS) orbits the earth with a velocity 7.6 km/s. How much energy is required to lift a payload o
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Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:
In our problem we have and
, so we can find the magnitude of the electric field:
The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to
Therefore, the magnitude of the force acting on the proton will be
And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
Answer:
Explanation:
Given that
The speed of the airplane ,v= 142 m/s
The speed of the air ,u = 30 m/s
Lets take angle make by airplane from east direction towards north direction is θ .
Now by using diagram ,we can say that
Now by putting the values in the above equation we get
Therefore the angle will be 12.19° .
