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stiv31 [10]
2 years ago
10

A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d

istance of 400 mm . Suppose a runner, moving at a constant speed, goes once around the track in 1 minmin 40 s.What is her centripetal acceleration during the turn at each end of the track?
Physics
1 answer:
lisabon 2012 [21]2 years ago
6 0

The centripetal acceleration a is 4.32 \times 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m,       Time = 100 s.

Computing the v = 0.4 m / 100 s

                         v = 4 \times 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

            centripetal acceleration a = v^2 / r

                                                        = (4 \times 10^-3)^2 / 0.037

                                                    a = 4.32 \times 10^-4 m/s^2.

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A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall
Alexandra [31]

Answers:

a) \theta_{2}=38\°

b) t=0.495 s

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

x=V_{o}cos\theta_{1} t_{1}   (1)

Where:

V_{o}=14.1 m/s is the initial speed  of snowball 1 (and snowball 2, as well)

\theta_{1}=52\° is the angle for snowball 1

t_{1} is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

y=0  is the final height of the  snowball 1

g=-9.8m/s^{2}  is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

x=V_{o}cos\theta_{2} t_{2}   (3)

Where:

\theta_{2} is the angle for snowball 2

t_{2} is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}   (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate t_{1} from (2):

0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}   (5)

t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}   (6)

Substituting (6) in (1):

x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})   (7)

Rewritting (7) and knowing sin(2\theta)=sen\theta cos\theta:

x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})   (8)

x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))   (9)

x=19.684 m   (10)  This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate t_{2} from (4) and substitute it on (3):

t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}   (11)

x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})   (12)

Rewritting (12):

x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})   (13)

Finding \theta_{2}:

2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})   (14)

2\theta_{2}=75.99\°  

\theta_{2}=37.99\° \approx 38\°  (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle (\theta_{2} < \theta_{1}).

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of t_{1} and t_{2} from (6) and (11), respectively:

t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}  

t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}   (16)

t_{1}=2.267 s   (17)

t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}  

t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}   (18)

t_{2}=1.771 s   (19)

Since snowball 1 was thrown before snowball 2, we have:

t_{1}-t=t_{2}   (20)

Finding the time difference t between both:

t=t_{1}-t_{2}   (21)

t=2.267 s - 1.771 s  

Finally:

t=0.495 s  

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A lightbulb has an effciency of 8.2%. How much light energy (not heat energy) is generated by the bulb every second if the buld
telo118 [61]
It really depends on what the bulb is being used for, since efficiency means how much of the output is USEFUL.
If the bulb is being used for light in a dark room, then it produces (8.2% x 21W) = 1.72 joules per second of light energy.
If the bulb is being used to keep a hamster cage or a fish tank warm, then the 8.2% is the useful part, and the light is the other (91.8% x 21W) = 19.28 joules per second.
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