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3 years ago

8

A plastic tube placed from the ventricles of the brain to the peritoneal cavity in order to continuously remove excess cerebrosp

inal fluid is called a ____________ shunt.

1 answer:

irina1246 [14]3 years ago

4 0

Answer:

ventriculoperitoneal

Explanation:

A VP or ventriculoperitoneal shunt is a medical equipment or a device which is used to relieve pressure from the brain that is caused by the accumulation of fluid in the brain. Ventriculoperitoneal shunting is a medical procedure that is primarily used to treat a condition known as hydrocephalus. This condition occurs when the more cerebrospinal fluid (CSF) gets collected in the ventricles of the brain than required.

Hence the answer is ---

ventriculoperitoneal

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A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente

inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

$F=\frac{mv^2}{r}$

Where, $m$ is the mass of the ball, $v$ is the speed and $r$ is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

$20.4 = \frac{3.1\times v^2}{2}$

and, $v^2 = 13.16$

Thus, $v = 3.63$ m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = $\frac{\text {Path length} }{\text {Speed}}$ = $\frac{12.56}{3.63}$ = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0

3 years ago

A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the

Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0

4 years ago

In this reaction diagram which part represents the doffrence in energy between the reactants and the products?

Annette [7]

Answer:

The correct answer is - option C. G.

Explanation:

In this reaction diagram, there is a representation of the reaction profile. The reaction profile shows the change that takes place during a reaction in the energy of reactants or substrate and products. In this profile, activation energy looks like a hump in the line, and the minimum energy required to initiate the reaction.

The overall energy of the reaction, including or excluding activation energy depends on the nature of the reaction if it is exothermic or endothermic. and products are represented by the G which shows the difference between the energy of the reactants and products.

3 0

3 years ago

Read 2 more answers

The capacitance of A conductor is affected by the presence of A second conductor that is uncharged and isolated electrically. Wh

Komok [63]

What happens is the potential value of the conductor decreases due to the presence of second conductor
as the capacitance is given by C = q/v
the value of v deceases as v-v1
thus the new capacitance is = C' = q/v-v1 thus the lowering of v increases the capacitance

5 0

3 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago