Answer:
Distance = 13.9 meters
Explanation:
Given the following data;
Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s
Decelerating speed = 3m/s
To find the distance travelled with this speed;
Distance = maximum speed/decelerating speed
Distance = 41.67/3
Distance = 13.9 meters
Therefore, the bus would travel a distance of 13.9 meters before stopping.
The answer is Homeostasis
To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,
![sin\theta = 1.22\frac{\lambda}{d}](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%201.22%5Cfrac%7B%5Clambda%7D%7Bd%7D)
Here,
= Wavelength
d= Diameter of aperture
= Angular resolution or diffraction angle
Our values are given as,
![\theta = 11\°](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2011%5C%C2%B0)
The frequency of the sound is ![f = 9100 Hz](https://tex.z-dn.net/?f=f%20%3D%209100%20Hz)
The speed of the sound is ![v = 343 m/s](https://tex.z-dn.net/?f=v%20%3D%20343%20m%2Fs)
The wavelength of the sound is
![\lambda = \frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bv%7D%7Bf%7D)
Here,
v = Velocity of the wave
f = Frequency
Replacing,
![\lambda = \frac{(343 m/s)}{(9100 Hz)}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B%28343%20m%2Fs%29%7D%7B%289100%20Hz%29%7D)
![\lambda = 0.0377 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.0377%20m)
The diffraction condition is then,
![sin\theta = 1.22\frac{\lambda}{d}](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%201.22%5Cfrac%7B%5Clambda%7D%7Bd%7D)
Replacing,
![sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}](https://tex.z-dn.net/?f=sin%2811%5C%C2%B0%29%20%3D%201.22%5Cfrac%7B%280.0377%20m%29%7D%7B%28d%29%7D)
d = 0.24 m
Therefore the diameter should be 0.24m
Answer:
Atomic size decreases from left to right in te periodic table.
Explanation:
With increase in effective nuclear charge without a corresponding increase in the number of shells across the period, electrostatic interaction between electrons and the nucleus increases with the net effect that atomic size decreases towards the right of the periodic table. For this reason, there are peaks at the larger alkali metals and valleys at the smaller noble gases.
Radians per second (rad/s)