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yan [13]
3 years ago
15

in a 4 kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kil

ometer in 6.3 minutes, and the final kilometer in 6 minutes. the average speed of the runner for the race is approximately
Physics
2 answers:
ohaa [14]3 years ago
5 0
To find the average of data collected, add all of the measurements: 5.9+6.2+6.3+6= 12.2

Then, divide the total amount by the number of data collected which is 4: 12.2/4= 3.05

The average speed of the runner of the race is approximately 3.05 km/min

Feel free to ask me any other questions you might have :)
Kruka [31]3 years ago
5 0

Answer:

9.83 km/h

Explanation:

Given: in a 4 kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6 minutes.

To Find:  the average speed of the runner for the race.

Solution:

In a 4 kilometer race,

time required to travel first kilometer = 5.9 \text{minutes}

time required to travel seoond kilometer = 6.2 \text{minutes}

time required to travel third kilometer = 6.3 \text{minutes}

time required to travel final kilometer = 6 \text{minutes}

we know that,

\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}

Total distance traveled = 4  \text{kilometer}

Total time taken = (5.9+6.2+6.3+6)

                           =  24.4 \text{minutes}

                           = \frac{24.4}{60}  \text{hours}

putting values,

\text{Average speed}=\frac{4}{\frac{24.4}{60}}

 \text{Average speed}=\frac{4\times60}{24.4}

                                            =9.83 \text{km}/ \text{h}

the average speed of the runner for the race is approximately 9.83 \text{km}/ \text{h}

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E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

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E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

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