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!!!
Answer:
Explanation:
When the box is on the ramp , component of its weight along the ramp
= mg sinθ
Friction force acting on it in upward direction
=μ mg cosθ
For sliding
μ mg cosθ < mg sinθ
μ cosθ < sinθ
.5 x cos35 < sin35
.41 < .57
So the box will slide
When sliding starts , kinetic friction acts
Net force in downward direction
mgsinθ - μ mg cosθ
acceleration
= gsinθ - μ g cosθ
= 5.62 - .3 x 9.8 x cos35
= 5.62 - 2.4
= 3.22 m /s²
I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.
Work = 20 N * 6 m * 1
Work = 120 Nm
Work = 120 joules
Hope this helps!
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