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3 years ago

How was dyslexia treated in the past and how is dyslexia treated today?

1 answer:

4 0

Dyslexia is treated using specific educational approaches and techniques, and the sooner the intervention begins, the better.

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a spring is used to launch a ball vertically into the air. the spring has a spring constant of 200N/m and is compressed by 5 cm.

zhannawk [14.2K]

Answer:

2.55 m

Explanation:

Elastic energy = gravitational energy

½ kx² = mgh

h = kx² / (2mg)

h = (200 N/m) (0.05 m)² / (2 × 0.010 kg × 9.8 m/s²)

h = 2.55 m

8 0

3 years ago

If two connected points objects pass through the same set of three points, the shapes created by each will be identical, regardl

djverab [1.8K]

Answer:

True

Explanation:

It Depends on the order in which each object was plotted, if two connected points objects pass through the same set of three points, the shapes created by each point may be different

7 0

3 years ago

Nerve signals in the body occur when a small voltage, called an action potential, is applied across the membrane of a cell. When

nasty-shy [4]

Answer: 23.56 nV

Explanation:

Assuming that we apply Ohm's Law to this situation, we know that under this condition, the current resultant, is proportional to the voltage applied, and that the proportionality constant, is called the resistance.

Now, we define electric current, as the passage of a number of charges over time.

In this case, we can say that each singly-ionized potassium ion carries the charge equivalent to the one electron, which is q = 1.6. 10⁻¹⁹ coulombs.

So, as we have 90,000 ions, we will have a total charge as follows:

Q =90.10³. 1.6.10⁻¹⁹ coulombs

Also, we are told that this charge will be moving for 1.1 msec, so we can find the current I as follows:

I = Q/t ⇒ I = 13.09. 10⁻¹² A.

If we know that R= 1.8.10⁹ Ω, we can determine V, applying Ohm's Law, as follows:

V = 13.09. 10⁻¹² A . 1.8.10⁹ Ω = 23.56 nV

3 0

3 years ago

Read 2 more answers

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

7 0

3 years ago

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

3 years ago