Answer:
They both tend to develop during the spring (March-June), reach peak intensity during the late autumn or winter (November-February), and then weaken during the spring or early summer (March-June)
What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
Read more about Electrostatics at; brainly.com/question/18108470
Answer: Tension = 53.6N
Explanation:
Given that
Height h = 1 m
Time t = 1.7 s.
Mass m = 5.1 kg
From the equation of the motion we can get the acceleration of the elevator:
h = X0+ V0t + at2/2;
Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus
a = 2h/t2 = 2 × 1/1.7^2
a = 0.69 m/s2.
Then we can find the tension in the cord by using the formula
T = mg + ma
= 5.1 (9.8 + 0.69)
= 5.1 × 10.5
= 53.6N
Answer: The answer is 333.3333 repeating
Explanation:
Divide the mass by the volume.
The magnitude and direction of the electric field at the position of this charge.3.33 N/C upward
An electric field is the electric force per unit of charge. It is assumed that the field's direction corresponds to the direction in which a positive test charge would experience force. The electric field is directed radially inward toward a negative point charge and radially outward from a positive charge.
Value of force F given = 10N
value of charge Q = 3 C
We know that E = F/Q
E = 10/3
= 3.33N
where charge is scalar quantity so the direction of force is the direction of electric field
Hence the magnitude and direction of the electric field at the position of this charge.3.33 N/C upward
Learn more about Electric field here
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