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tino4ka555 [31]
3 years ago
6

A car traveling 26 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is it's acceleration. ___m

/s^2
Physics
1 answer:
V125BC [204]3 years ago
6 0
To find acceleration you subtract the final velocity and subtract it from the initial velocity and you divide the difference by the total time.

so you take the final velocity which is 0 m/s because the car comes to a complete stop.

then you subtract that from the initial which is 26 m/s

in other words:

0m/s - 26 m/s = -26 m/s

then you divide it by the total time which is 13 s

-26 m/s / 13 s

so you get

-2 m/s^2
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Which material(s) listed below is an example of a persistent organic pollutant?
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When large amplitude sound vibrates the ear significantly, a _______ results.
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3 years ago
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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

in this case we'll use the average radius of earch which is 6,371 km, so we get:

r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

a=24.07 mm/s^{2}

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Using the division operation, the <em>number of tubes gotten by each patient </em>can be obtained by <em>dividing 20 by 5</em>. Hence, each patient will get 4 tubes of solution.

<u>Tubes of solution received</u> :

  • 10 a.m = 2 tubes
  • 12 p.m = 4 tubes
  • 3 p.m = 8 tubes
  • 6 p.m = 6 tubes

<u>Total tubes received</u> :

  • (2 tubes + 4 tubes + 8 tubes + 6 tubes) = 20 tubes

  • Number of patients = 5 patients

<u>Number of tubes of solution gotten by each patient</u> :

  • (Total tubes of solution ÷ number of patients)

Tubes of solution per patient = (20 ÷ 5) = 4 tubes

Therefore, each patient will get 4 tubes of solution.

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romanna [79]
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5 0
4 years ago
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