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3 years ago

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The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of t

he axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.3 m long, and the nerve impulse speed is 33 m/s, how long (in s) does it take for the nerve signal to travel this distance?

1 answer:

zysi [14]3 years ago

3 0

Answer:

Explanation:

Speed of electrical nerve signal = 33 m /s

Distance travelled = 1.3 m

time taken = distance / speed

= 1.3 / 33

= .039 s

= 39 ms ( millisecond ) .

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Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

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Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

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$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

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Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

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Part f)

If we throw at same speed so that it reach maximum height

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$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

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Answer:

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<h2>Velocity of 0.2 kg ball is 0.17 m/s</h2>

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