Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be = 1.9000000000000001 J
and the final kinetic energy from point B be = ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
Answer:
Er = 231.76 V/m, 27.23° to the left of E1
Explanation:
To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:
E1:
E2:
Keep in mind that the x component of electric field E2 is directed to the left.
∑x:
∑y:
The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.
or 27.23° to the left of E1.
Answer D
In alkali earth metals reacrivity increases from top to bottom (opposite of b)
This is because as you go down, the electron shells increase by 1 shell. The farther away a shell is from the nucleus, the higher its tendency to react.
D is true because the more reactive an alkali metal is, the more vigorous the reaction will be with water.
you have to substitiute volume of the ballon