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3 years ago

What does a refractory period prevent

1 answer:

5 0

Keister since the fence after I sent the van the refractory period prevents double counting the same event where is after I passed a van they prevent sensing the patient stimulus it’s after

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An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one

aalyn [17]

Answer:

20.78 m/s that we can approximate to option d (21 m/s)

Explanation:

The solution involves a lot of algebra and to be familiar with different convenient formulas for launching an object vertically under the action of gravity.

First you need to recall (or derive) the formula for the maximum height reached by an object with launches with initial velocity $v_0$ :

Maximum height = $\frac{{v_0}^2}{2g}$

Therefore one fourth of such height would be: $\frac{{v_0}^2}{8g}$

Second, find what would be the time needed to reach that height by solving for the time in the equation for the vertical position:

$y(t)=v_0*t-\frac{g}{2} t^2 \\\frac{{v_0}^2}{8g} = v_0*t-\frac{g}{2} t^2\\\frac{g}{2} t^2-v_0*t+\frac{{v_0}^2}{8g}=0$

And now, solve for t in the last equation using the quadratic formula to find the time needed for the object to reach that height (one fourth of the max height):

$t=\frac{v_0+/-\sqrt{{v_o}^2-4*\frac{g}{2}*\frac{{v_o}^2}{8g} } }{g} = \frac{v_0+/-\sqrt{{v_o}^2-\frac{{v_o}^2}{4} } }{g} =\frac{v_0+/-\sqrt{\frac{3{v_o}^2}{4} } }{g} = \\=\frac{v_0+/-v_0\sqrt{\frac{3}{4} } }{g} =\frac{v_0+/-v_0\frac{\sqrt{3}}{2} } {g}$

Next, use this expression for t in the equation for the velocity at any time t in the object's trajectory that comes from the definition of acceleration;

$v(t)=v_0-g*t$

Then for the time we just found, this new equation becomes:

$v=v_0-g(\frac{v_0+/-v_0\frac{\sqrt{3} }{2}}{g}) =v_0-v_0+/- v_0\frac{\sqrt{3} }{2} = +/- v_0 \frac{\sqrt{3} }{2}$

Now, using that the velocity at this height is 18 m/s, and solving for the unknown velocity $v_0$ , we get:

$v_0=\frac{18*2}{\sqrt{3} } =\frac{36}{\sqrt{3} }= 20.78\frac{m}{s}$

7 0

3 years ago

Why the efficiency of an engine cannot be 100%?

Marysya12 [62]

<span>No machine can operate at 100 percent efficiency because some of the energy input will always be used to overcome the force of gravity and the effects of friction and air resistance.</span>

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3 years ago

Read 2 more answers

Highest common factor of 12r and 10

Scilla [17]

I think the number 2, not sure

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3 years ago

Help please an observer at the North Pole sees the moon phase shown below. What Moon phase will be observed approximately one we

ziro4ka [17]

Answer:

it should be a new moon

Explanation:

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3 years ago

At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?

Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$$m / s^2$

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$$m/s^2$ is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}$

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0

2 years ago