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Anna007 [38]
3 years ago
12

A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of

29.5 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 24.4 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track
Physics
1 answer:
Leto [7]3 years ago
5 0

Answer: 2.27\ m/s^2

Explanation:

Given

Length of the race track L=200\ m

the radius of curvature of the track r=29.5\ m

time taken to run on track is t=24.4\ s

Speed of runner is

\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s

Centripetal acceleration is

\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2

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3 years ago
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

        B  =  \frac{\mu_o  * N  *  I}{2 r }

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

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3 0
3 years ago
What is one result of warming up your body’s muscles before exercising?
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Slightly raising your body temperature while increase oxygen and blood circulation throughout your body
7 0
3 years ago
you travel 4.0km east, 4.0km north, then 5.0km at 53.1 degrees north of west in a total of 5 hours. What is the magnitude and di
Sergio039 [100]

Answer:

Explanation:

We shall convert all the displacement in vector form .

i and j represents east and north respectively .

D₁ = 4 i

D₂ = 4 j

D₃ = - 5 cos 53.1 i + 5 sin 53.1 j

= -3i + 4 j

Total displacement = D₁ + D₂ + D₃

= 4i + 4 j - 3i + 4 j

= i + 8j

magnitude of displacement = √( 1² + 8² )

= 8.06 km

velocity = 8.06 / 5

= 1.61 km / h

Direction from x axis in anticlockwise direction .

Tanθ =  8 / 1 = 8

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7 0
3 years ago
The quartz crystal in a digital watch has a frequency of 32. 8 khz. what is its period of oscillation?
strojnjashka [21]

The time period of oscillation is 3.04×10

what is time period?

A time period (denoted by 'T'' ) is the time taken for one complete cycle of vibration to pass a given point.[1] As the frequency of a wave increases, the time period of the wave decreases. The unit for time period is 'seconds'.

The related parameters are oscillation frequency and period.They are the opposite of the another.As a result the formula to be used is

frequency=1/time

given:

Frequency of oscillation=32.8khz=32800hz

therefore,

time period of oscillation=1/32800hz

time period of oscillation=3.04×10

learn more about oscillation from here: brainly.com/question/28167898

#SPJ4

3 0
1 year ago
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