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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Complete Question
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field
E⃗ 1 inside the cell at a distance of 3.05 μm from the center?
E⃗ 2 Just inside the surface of the cell
E⃗ 3 Just outside the surface of the cell
E⃗ 4 At a point outside the cell 3.05 μm from the surface
Answer:
E⃗ 1
0 V/m
E⃗ 2
0 V/m
E⃗ 3
E⃗ 4
Explanation:
From the question we are told that
The diameter is
The charge is
The permittivity of free space is
The distance considered is
Generally the electric field inside the cell at a distance of 3.05 μm from the center is
0 V/m
This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge
Generally the electric field just inside the surface of the cell is 0 V/m
This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge
Generally the electric field just outside the cell is mathematically represented as
Here is the coulomb constant with value
r is the radius of the sphere which is mathematically as
Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as
Here R is mathematically represented as
=>
So