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2 years ago

Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.

Physics

1 answer:

tamaranim1 [39]2 years ago

4 0

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

$s=ut+\frac{1}{2}*gt^{2} \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m$

Arrow misses the target by 5.843m ig the arrow us split horizontally

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2 years ago

"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f

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Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c = 25 ms = 25 × 10⁻³ s

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The final temperature $T_2$ = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume = 0.4 mV/K × 473 K = 189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2 + (189.2 - 109.2)( 1 - $\mathbf{e^{-t/c}}$ )mV

y(t) = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 × 189.2 mV

= 170.28 mV

170.28 mV = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

170.28 mV - 109.2 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

61.08 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

0.7635 mV = ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

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The error fraction = 0.1

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