Which of the following is an accurate description of synaptic pruning?

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Yuliya22 [3333]

Answer:

The horizontal distance d that the ball covers before it lands is 1.72 m.

Explanation:

Given,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

We need to determine the horizontal distance d the ball travels before landing.

We need to calculate the time

Using the equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

Next, we can find the ball's velocity

Using the kinetic energy formula

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

By applying the conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

We substitute the values into the equation

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

Next, we determine the horizontal distance d the ball travels before landing

Using the distance formula

$d =vt$

Where. d = distance

t = time

v = velocity

We substitute the values into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Thus, the horizontal distance d that the ball travels before it lands is 1.72 m.

8 0

2 months ago

What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?

Keith_Richards [3271]

Based on my findings, within a period of 2 hours, there are certain atoms remaining. N = N0 * 2^(-t/6.020) = N = N0 * 2^-0.33223 = 0.7943 N0 Thus, the quantity of atoms that undergo disintegration is N0 - N = N0 * (1 - 0.79430) = 0.2057 N0 This must equate to 15 mCi = 15 * 3.7 * 10^7 = 5.55 * 10^8 atoms N0 = 5.55 * 10^8 / 0.2057 = 2.698 * 10^9 atoms Consequently, 2.698 * 10^9 atoms represents the value of N0.

4 0

3 months ago

Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to

Sav [3153]

Arginine is classified as a basic amino acid since it has two amino groups alongside a single acid group. At a low pH level, all ionizable groups are protonated. As the pH rises slightly, the acid group loses its proton. When the pH increases further, one of the amino groups also loses a proton. At considerably high pH levels, none of the ionizable groups remain protonated.

Pkas

pka1 = 1.82 pka2 = 8.99 pka3 = 12.48 Thus, 9.20 is above the second pKa and below the third pKa. This indicates that the acid has already lost its proton, as has one of the amino groups, while the second amino group remains protonated. When an acid is not protonated, it carries a negative charge. An unprotonated amino group is neutral, whereas when protonated, the amino group bears a positive charge. Therefore, this amino acid exhibits one positive charge (from one of the amino groups) and one negative charge (from the acid), resulting in an overall neutral charge.

4 0

2 months ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

serg [3582]

Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Distance = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

The potential difference is calculated as

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference across the plates amounts to 10000 V

Area is determined by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$

4 0

2 months ago

The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. how slowly can the ball go through its top

Keith_Richards [3271]

A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack. centripetal force = weight of the ball m v^2 / r = m g v^2 / r = g v^2 = g r v = sqrt { g r } v = sqrt { (9.80~m/s^2) (0.7 m) } v = 2.62 m/s Thus, the minimum speed for the ball at the top position is 2.62 m/s.

6 0

3 months ago