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3 years ago

15

While in a park, you walk west for 52 m, then you walk 33.9° north of west for 44 m, and finally you walk due north for 25 m. Fi

nd the components of your final displacement, from your initial to final point, along the north and west directions. (a) displacement component due north (b) displacement component due west

1 answer:

vodka [1.7K]3 years ago

4 0

Answer:

(a) Total north component = (26.89 + 25) = 51.89 metres

(b) Total west component = (-34.83 + 52) = 17.17 metres.

Explanation:

Do northwest components first.

North component = (sin 33.9 x 44) =26.89 metres.

West component = (cos 33.9 x 44) = -34.83metres.

Total west component = (-34.83 + 52) = 17.17 metres.

Total north component = (26.89 + 25) = 51.89 metres

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(a) 3.58 km 45° south of east

The total displacement is given by:

$d=vt$

where

v is the average velocity

t is the time

The average velocity is:

v = 3.53 m/s

While we need to convert the time from minutes to seconds:

$t=169 min \cdot 60 s/min = 10140 s$

Therefore, the magnitude of the displacement is

$d=(3.53)(10140)=35794 m = 3.58 km$

And the direction is the same as the velocity, therefore 45° south of east.

(b) 5.53 m/s 90° south of east

The velocity of the air relative to the ground is

$v_a = 2.00 m/s$

and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is

v = 3.53 m/s

So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed ( $v_a$ ). Since these two vectors are in opposite direction, we have

$v= v'-v_a$

Therefore we find v', Allen's velocity relative to the air:

$v'=v+v_a = 3.53 + 2.00 = 5.53 m/s$

The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.

(c) 56.1 km at 90° south of east.

Since Allen's velocity relative to the air is

v' = 5.53 m/s

Then the displacement of Allen relative to the air will be given by

$d'=v't$

and substituting,

$d'=(5.53)(10140)=56074 m = 56.1 km$

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two ice skaters push each other, and move in opposite directions. the mass of one is 75 kg, and his speed is 2 m/s. if the mass

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momentum conservation

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4 years ago

What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H

Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 : $= [H^+]=0.01 mol/L$

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Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

$pH=-\log [H^+]$

The hydrogen ion concentration at pH is equal to 2 = [H^+]

$2=-\log [H^+]\\$

$[H^+]=10^{-2}M= 0.01 M=0.01 mol/L$

The hydrogen ion concentration at pH is equal to 6 = [H^+]

$6=-\log [H^+]\\\\$

$[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L$

Concentration of hydrogen ion at pH is equal to 2 = $[H^+]=0.01 mol/L$

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Moles of hydrogen ion in 0.009999 mol/L solution :

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There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

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