Let v = the running speed
After running at constant speed for 26 min, the distance traveled is
d = (v m/min)*(26 min) = 26v m
Because there are 1500 m to go, the distance traveled is
10000 - 1500 = 8500 m
The running speed is
v = (8500 m)/(26 min) = 326.9 m/min
In km/h, the speed is
v = (0.3269 km/min)*(60 min/h) = 19.6 km/h
Answer: The running speed is 19.6 km/h
Answer:
The current in primary coil is 2.08 A.
Explanation:
Given that,
Power = 500 W
Voltage = 16 kV
Number of turns = 80000
We need to calculate the number of current
Using formula of voltage
![\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%3D%5Cdfrac%7BV_%7B1%7D%7D%7BV_%7B2%7D%7D)
Put the value into the formula
![\dfrac{N_{1}}{80000}=\dfrac{120}{16\times10^{3}}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_%7B1%7D%7D%7B80000%7D%3D%5Cdfrac%7B120%7D%7B16%5Ctimes10%5E%7B3%7D%7D)
![N_{1}=\dfrac{240}{16\times10^{3}}\times80000](https://tex.z-dn.net/?f=N_%7B1%7D%3D%5Cdfrac%7B240%7D%7B16%5Ctimes10%5E%7B3%7D%7D%5Ctimes80000)
![N_{1}=1200](https://tex.z-dn.net/?f=N_%7B1%7D%3D1200)
We need to calculate the current in secondary coil
Using formula of current
![i=\dfrac{P}{V_{2}}](https://tex.z-dn.net/?f=i%3D%5Cdfrac%7BP%7D%7BV_%7B2%7D%7D)
![i_{2}=\dfrac{500}{16\times10^{3}}](https://tex.z-dn.net/?f=i_%7B2%7D%3D%5Cdfrac%7B500%7D%7B16%5Ctimes10%5E%7B3%7D%7D)
![i_{2}=0.03125\ A](https://tex.z-dn.net/?f=i_%7B2%7D%3D0.03125%5C%20A)
We need to calculate the current in primary coil
![\dfrac{I_{1}}{I_{2}}=\dfrac{N_{2}}{N_{1}}](https://tex.z-dn.net/?f=%5Cdfrac%7BI_%7B1%7D%7D%7BI_%7B2%7D%7D%3D%5Cdfrac%7BN_%7B2%7D%7D%7BN_%7B1%7D%7D)
Put the value into the formula
![\dfrac{i_{1}}{0.03125}=\dfrac{80000}{1200}](https://tex.z-dn.net/?f=%5Cdfrac%7Bi_%7B1%7D%7D%7B0.03125%7D%3D%5Cdfrac%7B80000%7D%7B1200%7D)
![i_{1}=\dfrac{80000}{1200}\times0.03125](https://tex.z-dn.net/?f=i_%7B1%7D%3D%5Cdfrac%7B80000%7D%7B1200%7D%5Ctimes0.03125)
![i_{1}=2.08\ A](https://tex.z-dn.net/?f=i_%7B1%7D%3D2.08%5C%20A)
Hence, The current in primary coil is 2.08 A.