What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizo
1 answer:
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
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Answer:
The magnitude of force is 1.86 N and the direction of force is towards the other wire.
Explanation:
Given:
Current flowing through each power line, I = 130 A
Distance between the two power lines, d = 40 cm = 0.4 m
Length of power lines, L = 220 m
The force exerted by the power lines on each other is given by the relation:
Substitute the suitable values in the above equation.
F = 1.86 N
Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.
Answer:
The volume is decreasing at 160 cm³/min
Explanation:
Given;
Boyle's law, PV = C
where;
P is pressure of the gas
V is volume of the gas
C is constant
Differentiate this equation using product rule:
Given;
(increasing pressure rate of the gas) = 40 kPa/min
V (volume of the gas) = 600 cm³
P (pressure of the gas) = 150 kPa
Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( );
(600 x 40) + (150 x ) = 0
Therefore, the volume is decreasing at 160 cm³/min