On the report evaluation, the teacher wrote, "Good idea, but you
1 answer:
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Answer:
1000 N
Explanation:
The magnitude of the electrostatic force between two charged object is given by
where
k is the Coulomb constant
q1, q2 is the magnitude of the two charges
r is the distance between the two objects
Moreover, the force is:
- Attractive if the two forces have opposite sign
- Repulsive if the two forces have same sign
In this problem:
are the two charges
r = 3000 m is their separation
Therefore, the electric force between the charges is:
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:
Length of third side = 3.97m
Explanation:
First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.
From A-B we have distance = 5.5m = Say it c
From B-C we have distance = 4.3m = Say it a
From A-C we have distance = ? = Say it b which we have to find out.
Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)
For our case it is:
b² = a² + c² - 2acCos(B) - Say it equation 1
From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.
There values are:
a = 4.3m; c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees
Now by putting the respectve values in equation 1 we have:
b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)
b² = 18.49 + 30.25 - 32.95
b² = 15.79
b = √15.79
b = 3.97m
Thus the length of third side is 3.97m.
PS: The picture of triangle is being attached for yours understanding.
