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
Sodium chloride, NaCl, is formed when a sodium atom transfers its electron to a chlorine atom. The difference in charge between the two atoms creates a(n) electrostatic attraction that bonds them together.
A)
It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:
Through Definition of Velocity, comes:
B)
Using the Velocity Hourly Equation in vertical direction, we have:
The angle of impact is given by:
If you notice any mistake in my english, please let me know, because i am not native.
Answer: The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another. Another way of stating this law of chemistry is to say the total energy of an isolated system.
Explanation:
