Answer: 1433.3 m/min
Explanation:
For 86 Km/h converted to a (m/min), convert kilometers to meters, and hour to minutes
So, 86 Km/h means 86 kilometers per 1 hour
- If 1 kilometer = 1000 metres
86 kilometers = 86 x 1000 = 86,000m
- If 1 hour = 60 minute
1 hour = 60 minutes
In m/min: (86,000m / 60 minute)
= 1433.3 m/min
Thus, 86 Km/h convert to 1433.3 m/min
This problem is to let you practice using Newton's second law of motion:
Force = (mass) x (acceleration)
-- The airplane's mass when it takes off (before it burns any of its load of fuel) is 320,000 kg.
-- The force available is (240,000 N/per engine) x (4 engines) = 960,000 N.
-- Now you know ' F ' and ' mass '. Use Newton's second law of motion to calculate the plane's acceleration.
Answer:
21 m/s.
Explanation:
The computation of the wind velocity is shown below:
But before that, we need to find out the angles between the vectors
53° - 35° = 18°
Now we have to sqaure it i.e given below
v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°
v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951
v^2 = 440.6
v = √440.6
v = 20.99
≈ 21 m/s
Hence, The wind velocity is 21 m/s.
The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.
If it accelerates at 20 m/s² during the hit, then
Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .
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!!!
