To answer the following questions for this specific problem:
a. 11.48 secs
b. Vp = a*t*3.6 =
3*11.48*3.6 = 124.0 km/h
<span>c. 9.1 secs. </span>
I am hoping that this answer has satisfied your query about
and it will be able to help you.
Answer:
C. a full outer shell of valence electrons.
Explanation:
The noble gases has a full outer shell so they don't have to react with other elements to gain or loose electrons (to have a full outer shell and be stable).
Answer:
a) 5 N b) 225 N c) 5 N
Explanation:
a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:
F₁₂ = K Q₁ Q₂ / r₁₂²
So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N
b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.
So, we will have F₁₂ = 9. 25 N = 225 N
c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:
F₁₂ = 25 N .1/5 = 5 N
When t=2, the ball has fallen d(2) = 16 (2²) = 64 feet .
When t=5, the ball has fallen d(5) = 16 (5²) = 400 feet .
Distance fallen from t=2 until t=5 is (400 - 64) = 336 feet.
Time period between t=2 until t=5 is (5 - 2) = 3 seconds.
Average speed of the ball from t=2 until t=5 is
(distance covered) / (time to cover the distance)
= 336 feet / 3 seconds = 112 feet per second.
That's what choice-C says.
The correct answer would be 1.35m/s sw.
