Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)
Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
Let point A be 0.0 miles (first city)
Let point B be 160.5 miles (first city to second city)
Let point C be 28.5 miles (first city to mail stop)
Take C – A = C [28.5 - 0.0 = 28.5] (This checks the distance between city 1 and Mail stop)
Then Take B – C = Distance from the first city to the second city [160.5 - 28.5 = 132 Miles]
Answer: The Mail stop is 132 miles from the Second City.
Answer:
0.125m/s^2
Explanation:
20-10=10
10 divided by 80=0.125m/s^2
Answer:
The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s
= 16.125 Pa
Explanation:
The Bernoulli's equation is essentially a law of conservation of energy.
It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.
For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.
We also assume that the initial velocity of wind is 0 m/s.
This calculation is presented in the attached images to this solution.
Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.
The density is obtained to be 1.29 kg/m³.
Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.
We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.
Hope this Helps!!!