19) A navy cargo plane drops drops bundles of shelters and mosquito nets onto a field in rural
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Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:
Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution
Answer:
a = - 50 [m/s²]
Explanation:
To solve this problem we simply have to replace the values supplied in the given equation.
Vf = final velocity = 0.5 [m/s]
Vi = initial velocity = 10 [m/s]
s = distance = 100 [m]
a = acceleration [m/s²]
Now replacing we have:
The negative sign of acceleration means that the ship slows down its velocity in order to land.
Electromagnet is in form of solenoid
and the magnetic field due to solenoid is given as
here
i = current in the loop
so when we increase the current in electromagnet the magnetic field of the solenoid will increase
this will increase the strength of the electromagnet
so the answer would be
<em>INCREASE</em>