Answer:
im not the best at explaining but i say the second graph is showing a faster moving object because the line is inclining vs the other graphs line which isnt going that much up but staying just at a consistent paste
Explanation:
im sorry if im wrong
The portion of the flux leaves the curved surface of the cylinder is 60%.
<h3 /><h3>What are electrons?</h3>
The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.
If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.
If 20% of the flux leave from one end, then another 20% will leave from another end.
So, the net flux through curved surface is
100 -20 -20 = 60%
Thus, the total flux leaves the curved surface of the cylinder is 60%
Learn more about electrons.
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<h2>
Answer: Light reflects when it hits a surface.</h2>
Explanation:
It is now clear that light behaves as a wave and as a particle. It should be noted that the first to propose the <u>corpuscular theory </u>of light was <u>Issac Newton</u>, while the <u>wave theory</u> was initially proposed by <u>Christian Huygens</u>, who was contemporaneous with Newton.
Now, focusing on the corpuscular theory, <u>Newton proposed that light is composed of tiny massless particles, traveling in a straight line and at high speed.</u> In addition, he used the reflection phenomenon of the of light to show that it behaved like particles that when hitting a mirror were reflected by a perfectly elastic collision.
<span>Even in space, there is still presence of gravity. The
cause of weightlessness is not how far above the earth the space shuttle is but
rather how fast it is travelling. The shuttle is in free fall causing
weightlessness, but it is travelling fast enough to miss the earth as it falls.
Similarly, the airplane could also provide weightlessness if it went free fall
as well. However, that ends as the plane hits the ground. </span>
Answer:
Work done = 4584.9 J
Explanation:
given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C
Solution:
Formula for the potential difference at the center of the circle
P.E = K × q1 q2 /r (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)
P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C / 0.20 m
P.E = 4584.9 J = Work done