Answer:
9.44 eV
Explanation:
q = Amount of charge = 86.3 μC = 86.3 x 10⁻⁶ C (Since 1 μC = 10⁻⁶ C)
V = Electric potential of the charge = 17.5 Volts
U = Amount of electric potential energy carried by charge
Amount of electric potential energy carried by charge is given as
U = q V
Inserting the values
U = (86.3 x 10⁻⁶) (17.5)
U = 0.00151 J
U = 
U = 9.44 eV
KE = Kinetic energy the charge could have
Using conservation of energy
KE = U
KE = 9.44 eV
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We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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