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
The average distance from the Sun to Neptune is about 2.795 billion miles.
That's roughly 0.00048 of a light year .
m = Mass of the refrigerator to be moved to third floor = 136 kg
g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²
h = Height to which the refrigerator is moved = 8 m
W = Work done in lifting the object
Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence
Work done = Gravitation potential energy of refrigerator
W = m g h
inserting the values
W = (136) (9.8) (8)
W = 10662.4 J
Answer:
C. 14400
Explanation:
40min × 6 periods = 240mins total.
1 minute = 60 seconds.
240 minutes × 60 = 14400 seconds.
B.
technically it would depend if the resistors were in series or parallel but B is the answer.
