Answer:
Explanation:
Average speed = Total distance / Total time.
100 km/hr
r = 100 km / hr
t = 6 hours
d = 6 * 100 = 600 km
120 km / hr
r = 120 km / hr
t = 5 hour
d = 120 * 5
d = 600 km
Total distance = 600 + 600 = 1200 km
Total time = 5 hour + 6 hours = 11 hours.
Average speed = 1200 km / 11 hours = 109.1
Answer:
The weather will clear up and get sunnier.
Explanation:As weather forecasters monitor air pressure, falling barometer measurements can signal that bad weather is on the way. In general, if a low pressure system is on its way, be prepared for warmer weather with storms and rain. If a high pressure system is coming, you can expect clear skies and cooler temperatures.
hope this helped:)
Brainliest?
Answer:
the correct answer is C
Explanation:
When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.
Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.
Applying this to our case if there is
10 boulder on the chart
in reality there are #_boulder = 10 30
#_boulder = 300 boulder
so the correct answer is C
Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :
![v=f_1\lambda_1](https://tex.z-dn.net/?f=v%3Df_1%5Clambda_1)
Wavelength for f = 45 Hz is,
![\lambda_1=\dfrac{v}{f_1}](https://tex.z-dn.net/?f=%5Clambda_1%3D%5Cdfrac%7Bv%7D%7Bf_1%7D)
![\lambda_1=\dfrac{343}{45}=7.62\ m](https://tex.z-dn.net/?f=%5Clambda_1%3D%5Cdfrac%7B343%7D%7B45%7D%3D7.62%5C%20m)
Wavelength for f = 375 Hz is,
![\lambda_2=\dfrac{v}{f_2}](https://tex.z-dn.net/?f=%5Clambda_2%3D%5Cdfrac%7Bv%7D%7Bf_2%7D)
![\lambda_2=\dfrac{343}{375}=0.914\ m/s](https://tex.z-dn.net/?f=%5Clambda_2%3D%5Cdfrac%7B343%7D%7B375%7D%3D0.914%5C%20m%2Fs)
So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.