The ice melts a liquid the molecules spread out and when a liquid changes to a gas the molecules spread out more.
Answer:
The banking angle of the road is 15.8⁰.
Explanation:
Given;
radius of the curved road, r = 52 m
velocity of the traveling car, v = 12 m/s
acceleration due to gravity, g = 9.8 m/s²
let the banking angle of the road = θ
The banking angle of the road is calculated as;
![\theta = tan^{-1}(\frac{(12)^2}{52 \times 9.8} )\\\\\theta = 15.8^0 \\\\](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%28%5Cfrac%7B%2812%29%5E2%7D%7B52%20%5Ctimes%209.8%7D%20%29%5C%5C%5C%5C%5Ctheta%20%3D%2015.8%5E0%20%5C%5C%5C%5C)
Therefore, the banking angle of the road is 15.8⁰.
Answer:
Explanation:
By conservation of energy, speed of the ball going up = speed of ball coming down with the ball stops at the top.
Because the gravity acceleration is constant, by symmetry, half of total time, 6/2 = 3s, is for going up and the last 3s for coming down.
Consider the last 3s when the ball drops from top to bottom, the initial velocity = 0 and acceleration = 10m/s^2
distance traveled = initial velocity * time + 1/2 * acceleration * time^2
= 0*3 + 1/2*10*3^2
= 5*9
= 45m
So maximum height of the ball is 45m.
Answer:
38.35 bar
Explanation:
We are given that
Temperature=T=36.6 degree Celsius=36.6+273=309.6 K
Given mass of glucose=9.18 g
Molar mass of glucose(
=180 g
Mass of c=12 g,mass of hydrogen=1 g, mass of O=16 g
Volume of solution=34.2 mL
Molarity of solution=![\frac{given\;mass}{molar\;mass\times volume}\times 1000](https://tex.z-dn.net/?f=%5Cfrac%7Bgiven%5C%3Bmass%7D%7Bmolar%5C%3Bmass%5Ctimes%20volume%7D%5Ctimes%201000)
Where volume (in mL)
Molarity of solution=![\frac{9.18}{180\times 34.2}\times 1000=1.49 M](https://tex.z-dn.net/?f=%5Cfrac%7B9.18%7D%7B180%5Ctimes%2034.2%7D%5Ctimes%201000%3D1.49%20M)
We know that
Osmotic pressure=![\pi=MRT](https://tex.z-dn.net/?f=%5Cpi%3DMRT)
Where M=Molarity of solution
R=Constant=0.08314 Lbar/mol k
T=Temperature in kelvin
Using the formula
![\pi=1.49\times 0.08314\times 309.6=38.35 bar](https://tex.z-dn.net/?f=%5Cpi%3D1.49%5Ctimes%200.08314%5Ctimes%20309.6%3D38.35%20bar)
Hence, the osmotic pressure=38.35 bar
Answer:
Explanation: follow the derivation in attachment