Answer:
The turning points are those instants, moments or situations that happen in an absolutely unexpected way, as a result of which your life changes ... and nothing is the same as before.
So your finding acceleration first which is 30m/s divides by 6 seconds equals 5m/s^s and then multiply that by 1,400 kg and you have net force which is 7,000N
Answer: 539.4 N
Explanation:
Let's begin by explaining that Coulomb's Law establishes the following:
"The electrostatic force 
between two point charges 
and 
is proportional to the product of the charges and inversely proportional to the square of the distance 
that separates them, and has the direction of the line that joins them"
What is written above is expressed mathematically as follows:
(1)
Where:
is the electrostatic force
is the Coulomb's constant
and are the electric charges
is the separation distance between the charges
Then:
(2)
Isolating and :
(3)
Now, if we keep the same charges but we decrease the distance to 
, (1) is rewritten as:
(4)
Then, the new electrostatic force will be:
(5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.
True, They contain old stars and posses little gas or dust
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
