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
Answer:
5.714 hours / day
Explanation:
<u>Calculate the hours used in that week </u>
120000 / 3000 = 120 / 3 = 40 hours a week
<u>Calculate the amount it is used in one day</u>
40 / 7 = 5.71428571 hours or 5.714 hours/day
Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
The correct answer to the question is : D) Be moving at a constant velocity.
EXPLANATION:
As per Newton's first laws of motion, every body continues to be at state of rest or of uniform motion in a straight line unless and until it is compelled by some external unbalanced forces acting on it.
Hence, it is the unbalanced force which changes the state of rest or motion of a body. Balanced force is responsible for keeping the body to be either in static equilibrium or in dynamic equilibrium.
As per the options given in the question, the last one is true for an object under balanced forces.
Answer:
my butt
Explanation:
2x2 by the power of 10 divide that 98