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3 years ago

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An 18 month-old child is hypotonic and cannot fully extend the neck and trunk against gravity. Which position would be MOST EFFE

CTIVE to place the child in to facilitate antigravity movement and promote hand usage?
a. Fully upright in a standing frame
b. Prone on a 45 degree wedge positioned on the floor
c. Seated upright on a large therapy ball
d. Straddle-sit position on a bolster

1 answer:

Flura [38]3 years ago

5 0

Answer:

b. Prone on 45 degree wedge positioned on the floor.

Explanation:

According to the pediatric therapists the most effective position to place the child in to facilitate anti gravity movement and promote hand usage is to prone on 45 degree wedge positioned on the floor.

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A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to

Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

$\tau = I\alpha$

Where $\tau$ is the torque

$I$ is the moment of inertia

$\alpha$ is the angular acceleration

But, the angular acceleration is given by

$\alpha = \frac{\omega}{t}$

Where $\omega$ is the angular speed

and $t$ is time

Then, we can write that

$\tau = \frac{I\omega}{t}$

Hence,

$\omega = \frac{\tau t}{I}$

Now, to determine the angular speed, we first determine the Torque $\tau$ and the moment of inertia $I$ .

Here, The torque is given by,

$\tau = rF$

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ $\tau = 3.00 \times 195$

$\tau = 585$ Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

$I = \frac{1}{2}MR^{2}$

Where M is the mass and

R is the radius

∴ $I = \frac{1}{2} \times 325 \times (3.00)^{2}$

$I = 1462.5$ kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

$\omega = \frac{\tau t}{I}$

$\omega = \frac{585 \times 2.05}{1462.5}$

$\omega = 0.82$ rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

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A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic

vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 $m^{2}$ </h3>

Explanation:

Given :

No. of turns in the first coil $N_{1} = 41$

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Area of first coil $A_{1} = 0.074 m^{2}$

According to the law of electromagnetic induction,

Induced emf = $-N \frac{d \phi}{dt}$

Where $\phi =$ magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

$-\frac{N_{1} d\phi _{1} }{dt_{1} } = -\frac{N_{2} d\phi _{2} }{dt_{2} }$

$\frac{N_{1} A_{1} dB_{1} }{dt_{1} } = \frac{N_{2} A_{2} dB_{2} }{dt_{2} }$

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$A_{2} = 0.0246 = 0.025 m^{2}$

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