When birds remain airborne and moving without flapping their wings, on what are they gliding?
2 answers:
When birds remain airborne and moving without flapping their wings, they are gliding downward with maintaining thrust
<h3>Further explanation </h3>Birds are lightweight, because they have got the capability to clasp the wind in their wings. When the birds remain in the air but not flapping their wings, they are gliding across the wind. Birds have their wings open, so it makes them lighter across their whole body, and then it slowly glide.
They are gliding on updrafts or rising currents of air. Air close to the earth is heated by the sun and this causes it to rise. Warm air is less dense than cold air so it will rise. Birds seem to glide because the gravity force acting on the animal as it flies is equal by the rising column of air. This is called soaring. The definition of soaring is: rising very quickly to a high level. For example: A soaring California Condor spreads its primary feathers, so each acts as a small, and high-aspect-ratio wing. This reduces turbulence at the wingtips and lowers the stall speed
Birds also use the thermal uprise where the rising columns of air caused by the uneven heating of Earth's surface from solar radiation.
<h3>Learn more</h3>- Learn more about birds brainly.com/question/1595848
Grade: 9
Subject: physics
Chapter: birds
Keywords: physics, birds
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Answer:
a)
b)
Explanation:
In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).
After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:
We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:
When solving for the normal force we get:
we know that
W=mg
and
so after substituting we get that
N=F sin θ +mg
We also know that the kinetic friction is defined to be:
so we can find the kinetic friction by substituting for N, so we get:
Now we can find the sum of forces in x:
so after analyzing the diagram we can build our sum of forces to be:
we know that:
so we can substitute the equations we already have in the sum of forces on x so we get:
so now we can solve for the force, we start by distributing so we get:
we add to both sides so we get:
Nos we factor F so we get:
and now we divide both sides of the equation into so we get:
which is our answer to part a.
Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:
and
F cos θ = f
when substituting one into the other we get:
which can be solved for the static friction coefficient so we get:
which is the answer to part b.
