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2 years ago

Explain how the forces need to change so the aeroplane can land

1 answer:

7 0

When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.

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For your senior project, you are designing a Geiger tube for detecting radiation in the nuclear physics laboratory. This instrum

Paha777 [63]

Answer:

Maximum linear charge density = 84.14 nC/m

Explanation:

Looking at this question, The electric field of a line charge of infinite length is given by : Er = (1/(2πεo)) x (λ/r)

r = the distance from the center of the line of charge

λ = the linear charge density of the wire.

Now looking at the equatiom, due to the fact that Er varies inveresely with r, its maximum value will occur at the surface of the wire where r = R, the radius of the wire:

And so, Emax = (1/(2πεo)) x (λ/R)

Let's make λ the subject of the equation and we get;

λ = 2πεo(REmax)

From the question, R = 0.55/2 = 0.275cm

Also, Emax = 5.50 × 10^(6) N/C

Let's take the value of the electric constant to be εo = 8.854 x 10^(-9) C^(2) / Nm^2

R = 0.275mm = 0.000275m

Plugging these values into the equation, we get;

λ = 2π x 8.854 x 10^(-12) x 0.000275 x 5.50 × 10^(6) = 84.14 nC/m

4 0

2 years ago

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

2 years ago

which statements about velocity are true? check all that apply. a. for velocity, you must have a number, a unit, and a direction

satela [25.4K]

a). for velocity, you must have a number, a unit, and a direction.
Yes. This one isn't bad. The 'number' and the 'unit' are the speed.

b). the si units for velocity are miles per hour.
No. That's silly.
'miles' is not an SI unit, and 'miles per hour'
is only a speed, not a velocity.

c). the symbol for velocity is .
You can use any symbol you want for velocity, as long as
you make its meaning very clear, so that everybody knows
what symbol you're using for velocity.
But this choice-c is still wrong, because either it's incomplete,
or else it's using 'space' for velocity, which is a very poor symbol.

d). to calculate velocity, divide the displacement by time.
Yes, that's OK, but you have to remember that the displacement
has a direction, and so does the velocity.

3 0

2 years ago

A balloon is ascending at 12.4m/s at a height of 81.3m above the ground when a package is dropped. a) How long did it take to re

tankabanditka [31]

Answer:

3secs

Explanation:

Given the following parameters

height H= 81.3m

Velocity v = 12.4m/s

Required

Time it take to reach the ground