Your answer is cumulonimbus clouds
Answer:
14.2 m/s
Explanation:
Given data:
Speed of the stream, v₁ = 7.1 m/s
let the cross section area at initial point be A₁
now area at the second point, A₂ = (1/2)A₁ = 0.5A₁
now, from the continuity equation, we have
A₁v₁ = A₂v₂
where, v₂ is the velocity at the narrowed portion
thus, on substituting the values, we get
A₁ × 7.1 = 0.5A₁ × v₂
or
v₂ = 14.2 m/s
Answer:
The value is 
Explanation:
From the question we are told that
The initial pressure is
The initial temperature is ![T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K](https://tex.z-dn.net/?f=T_1%20%3D%20%2050%20%5C%20F%20%3D%20%2850%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D%20283%20%20%5C%20%20K)
The final temperature is ![T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K](https://tex.z-dn.net/?f=T_2%20%3D%20%20320%20%5C%20F%20%3D%20%28320%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D433%20%20%5C%20%20K)
Generally the equation for adiabatic process is mathematically represented as

=> 
Generally for a monoatomic gas 
So
![14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }](https://tex.z-dn.net/?f=14%20%2A%20283%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D%20%3DP_2%20%2A%20433%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D)
=> 
=> 
Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction


Put the value into the formula


Hence, The velocity of the student has after throwing the book is 0.0345 m/s.
Answer:
The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.