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2 years ago

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Most split-brain surgeries are preformed on ___________. A. neurology patients B. intractable epilepsy patients C. abnormal psyc

h patients D. college students

2 answers:

Alexandra [31]2 years ago

7 0

B. Intractable epilepsy patients

ivanzaharov [21]2 years ago

6 0

Answer: B. intractable epilepsy patients

Explanation:

Most split-brain surgeries are performed on intractable epilepsy patients. The corpus callosotomy is a treatment procedure in which the corpus callosum of the brain is cut so as the prevent the epileptic activity in the two hemispheres of the brain.

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In a nuclear reactor, the control rods are: A.made of graphite B.used to keep the reactor cool C.used to absorb free neutrons D.

algol13

I think its E. the control rods are used to control <span>the fission rate of uranium and plutonium. Hope this helps!! </span>

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3 years ago

Read 2 more answers

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\theta-\theta_{o} = 435\,rad$ , the angular deceleration is:

$\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}$

$\alpha = -8.780\,\frac{rad}{s^{2}}$

Now, the time interval of the Deceleration Phase is obtained from this formula:

$\omega = \omega_{o} + \alpha \cdot t$

$t = \frac{\omega - \omega_{o}}{\alpha}$

If $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\alpha = -8.780\,\frac{rad}{s^{2}}$ , the time interval is:

$t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }$

$t = 9.954\,s$

The total time needed for the grinding wheel before stopping is:

$t_{T} = 2.40\,s + 9.954\,s$

$t_{T} = 12.354\,s$

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0

3 years ago

The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov

son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The eyepiece focal length is $f_e = 0.027 \ m$

Explanation:

From the question we are told that

The focal length is $f_o = 5.5 \ mm = -0.0055 \ m$

This negative sign shows the the microscope is diverging light

The angular magnification is $m = 300$

The distance between the objective and the eyepieces lenses is $Z = 19 \ cm = 0.19 \ m$

Generally the magnification is mathematically represented as

$m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]$

Where $f_e$ is the eyepiece focal length of the microscope

Now making $f_e$ the subject of the formula

$f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }$

substituting values

$f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }$

$f_e = 0.027 \ m$

5 0

2 years ago

Q1. In a 100m track event, the time taken by four runners is 10 sec., 10.2 sec., 10.4 sec and 10.6 sec. Find the ratio of their

Anna35 [415]

Distance (s)=100m
time
T₁=10
T₂=10.2
T₃=10.4
T₄=10.6
by v= $\frac{s}{t}$
we get
V₁= $\frac{100}{10}$
V₁=10m/s

V₂= $\frac{100}{10.2}$
V₂=9.8m/s

V₃= $\frac{100}{10.4}$
V₃=9.61 m/s

V₄= $\frac{100}{10.6}$
V₄=9.43 m/s

V₁:V₂:V₃:V₄ = 10 : 9.8 : 9.61 : 9.43

8 0

2 years ago

A 512Hz tuning fork is used with a resonating column to determine the velocity of sound in water. If the spacing between resonan

EastWind [94]

Answer:

v= 1495.04 m/s

Explanation:

The formula for velocity of sound is given by ;

v= fλ --------where

v= velocity of sound

f= frequency of turning fork

λ = wavelength

However,

Δ L = 1/2 λ ------where Δ L is spacing between resonances.

1.46 = 1/2 λ

1.46 * 2 = λ

2.92 m = λ

v= fλ

v= 512 * 2.92

v= 1495.04 m/s

6 0

2 years ago