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3 years ago

What were the key factors that made the wright brothers’ wind tunnel a useful tool for wing design?

Physics

2 answers:

Anna71 [15]3 years ago

7 0

Answer:

Balancing equipment.

Explanation:

The wind tunnel made by Wright's brother was made up of a wooden box with glasses on it to see the inner instruments work.

The key factor that made the Wright's brother wind tunnel a useful tool for wind design is the balancing equipment used in it. It was designed to measure lift and drag. Old hacksaw blades and bicycle spokes were used to make the balances which measure the forces of lift and drag action on a wing.

IrinaK [193]3 years ago

5 0

The Wright brothers built a wind tunnel<span> and developed model-testing </span>techniques<span> including a </span>balance to more accurately determine the lift and drag of their aircraft. I hope my answer has come to your help. God bless and have a nice day ahead!

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An airplane increases its velocity from 60m/s to 80m/s in a time of 10s.

Elenna [48]

$\huge\fbox{Answer ☘}$

$acceleration \: = \frac{v - u}{t} \\ = > \frac{80 - 60}{10} \\ \\ = > \frac{20}{10} \\ \\ = > 2m \: s {}^{ - 2}$

hope helpful ~

4 0

2 years ago

An engine performs 6400 j of work on a motorbike the motorbike and the rider have a combined mass of 200 kg if the bike started

spin [16.1K]

The work done by the engine is converted into kinetic energy of the motorbike+rider system:

$W=K=\frac{1}{2}mv^2$

where

W=6400 J is the work

m=200 kg is the mass of the system

v is the speed acquired by the motorbike

Rearranging the equation and substituting the numbers in, we find

$v=\sqrt{\frac{2W}{m}} =\sqrt{\frac{2(6400 J)}{200 kg}} =8 m/s$

4 0

3 years ago

Saved

Anon25 [30]

Total distance moved by bead is 1.952 cm.

Explanation:

Let first consider all data that are given in question.

1. F = 8 N ...force acting on string

2. f = 2 Hz ...frequency of system

3. β = 4 cm = 0.04 m ...wavelength of wave formed due to vibration 4. A = 1 cm = 0.01 m ...Amplitude of vibration

Under certain conditions, waves can bounce back and forth through a particular region, effectively becoming stationary. These are called standing waves.

Here,it is due to vibration induced in spring due to tension induced in string

Standing wave equation is given by

$y = (x,t) = 2A * sinK x * cos (wt)$ ...(1)

Let first find, value of K, x, w, t

k = 2 * pi / beta ....(2)

Where β is wavelength in meters

K is wave number

k = 2 * pi / 0.01

k = 628.31 $m^{-1}$

now, let us find value of w

W = 2 x pi x f ....(3)

where f is frequency in hertz

W = 2 x pi x 2

W = 4 x pi

W = 0.08 $\frac{m}{s}$

$y = (x,t) = 2A * sinK x * cos (wt)$

now, let us find value of v that is wave speed

Notice that some x-positions of the resultant wave are always zero no matter what the phase relationship is. These positions are called nodes.

Finding the positions where the sine function equals zero provides the positions of the nodes.

In our case, and

K * x = pi

x = 0.04 / 2

x = 0.02

$y = (0.02,1) = 2(0.01) * sin pi * cos (12.5664 * 1)$

$y = (0.02,1) = 2(0.01) * -1 * cos 0.9761$

Y = 1.952 cm

Finally, when bead is at middle of the string, total distance after stretch covered is 1.952 cm.

3 0

3 years ago

Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

alukav5142 [94]

Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

4 0

3 years ago

What is the increase in pressure required to decrease volume of mercury by 0.001%

REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0

3 years ago