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1 year ago

12

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

1 answer:

tamaranim1 [39]1 year 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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Applying the law's theory and utilizing the equation of momentum ie. p=mv

Explanation:

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A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His

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Answer:

1) $v_2=23\ m.s^{-1}$ & $x_2=43\ m$ east of sign post

2) $x'=55\ m$ east of sign post

3) $x_n=205\ m$ east of the signpost.

4) $v_z=35\ m.s^{-1}$

Explanation:

Given:

position of motorcyclist on entering the city at the signpost, $x_0=0\ m$

time of observation after being at x=5m east of the signpost, $t_m=0\ s$

constant acceleration of the on entering the city, $a=4\ m.s^{-2}$

distance of the motorcyclist moments later after entering, $s_m=5\ m$

velocity of the motorcyclist moments later after entering, $u_m=15\ m.s^{-1}$

<u>Now the initial velocity on at the sign board:</u>

$u_m^2=u^2+2.a.x_m$

where:

$u=$ initial velocity of entering the city at the signpost

Putting respective values:

$15^2=u^2+2\times 4\times 5$

$u=13.6015\ m.s^{-1}$

1)

Position at time $t_2=2\ s$ sec.:

Using equation of motion,

$x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5$ because it has already covered 5m before that point

$x_2=15\times 2+0.5\times 4\times 2^2+5$

$x_2=43\ m$ east of sign post

Velocity at time $t_2=2\ s$ sec.:

$v_2=u_m+a.t_2$

$v_2=15+4\times 2$

$v_2=23\ m.s^{-1}$

2)

Position when the velocity is $v'=25\ m.s^{-1}$ :

using equation of motion,

$v'^2=u_m^2+2.a.x'+5$

$25^2=15^2+2\times 4\times x'+5$

$x'=55\ m$ east of sign post

3)

Given that:

acceleration be, $a_n=2\ m.s^{-2}$

time, $t_n=5\ s$

Position after the new acceleration and the new given time:

using equation of motion,

$x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5$

$x_n=15\times 5+0.5\times 2\times 5^2+5$

$x_n=205\ m$ east of the signpost.

4)

now time of observation, $t_z=5\ s$

$v_z=u_m+a.t_z$

$v_z=15+4\times 5$

$v_z=35\ m.s^{-1}$

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Answer:

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and The jnfusion rate = 41.67 gitt/hr

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10 gtts/ml × 50mg/6 × 30/60

Infusion rate = 15000/360

= 41.67 gitt/hr

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