The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
Learn more about electric field here: brainly.com/question/14372859
#SPJ1
You haven't said how much power the stereo uses. It matters !
Whatever that number is, the maximum hours per month is
(3460) divided by (the # of watts the stereo uses when it's playing) .
Answer:
Explanation:
v= u + at
v is final velocity , u is initial velocity . a is acceleration and t is time
Initial velocity u = 0 . Putting the given values in the equation
v = 0 + g sin 18 x 3.5
= 10.6 m /s
Answer:
112.23 m
Explanation:
Displacement is the final position minus the initial position.
Δx = x − x₀
Δx = 100.1 m − (-12.13 m)
Δx = 112.23 m
The area of the circle with radius r is
A = πr²
The rate of change of area with respect to time is
The rate of change of the radius is given as
Therefore
When r = 10 ft, obtain
Answer: - 40π ft²/s (or - 127.5 ft²/s)
