During __________, frequently used synaptic connections in the brain strengthen while rarely used connections tend to thin.

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

3 years ago

What goes through evaporation

Olegator [25]

Water goes through evaporation.

6 0

3 years ago

Read 2 more answers

You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin

Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

r = Re + 250

We substitute

W = G m M / r²

W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

W = G m M / Re² (1+ 250 / Re)²

(1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

(1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

W = Wo (1 - 7.85 10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0

3 years ago

Read 2 more answers

If you walk 3 kilometers in 30 minutes , what is the average speed in kilometers per hour?

Pepsi [2]

Answer:

6 km/h

Explanation:

V avg = ∆x/∆t = 3km / 30 min ×(60min/1h) = 3 km× 2 /h = 6 km/h

4 0

2 years ago

A car has uniformly accelerated from rest to a speed of 25m/s after traveling 75m. What is its acceleration in m/s^2

Roman55 [17]

<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’): </em></h2><h2><em> </em></h2><h2><em>The car’s speed is 25m/s </em></h2><h2><em>The distance travelled is 75m </em></h2><h2><em>Then we have the formulas for speed and distance: </em></h2><h2><em> </em></h2><h2><em>v = a x t -> 25 = a x t </em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2 </em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say: </em></h2><h2><em> </em></h2><h2><em>t = 25 / a </em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a </em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1): </em></h2><h2><em> </em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2 </em></h2><h2><em>t^2 = 150 / a </em></h2><h2><em>Since t has the same value for both truths we can say: </em></h2><h2><em> </em></h2><h2><em>625 / a^2 = 150 / a </em></h2><h2><em> </em></h2><h2><em>Thus multiply both sides with a^2: </em></h2><h2><em> </em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17 </em></h2><h2><em> </em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>

4 0

3 years ago