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4 years ago

True or false? the x axis is the horizontal axis and the y axis is the vertical axis

Physics

2 answers:

Sever21 [200]4 years ago

4 0

True;the x axis is the hor<span>izontal axis and the y axis is the vertical axis</span>

rosijanka [135]4 years ago

3 0

True
....................

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A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

slava [35]

Answer: the angular frequency is 2.31 rad/s

Explanation:

The data we have is:

Radial acceleration A = 27.9 m/s^2

Beam length r = 5.21m

The radial acceleration is equal to the velocity square divided the radius of the circle (the lenght of the beam in this case)

And we can write the velocity as:

v = w*r where r is the radius of the circle, and w is the angular frequency.

w = 2pi*f

where f is the "normal" frequency.

So we have:

A = (v^2)/r = (r*w)^2/r = r*w^2

We can replace the values and find w.

27.9m/s^2 = 5.21m*w^2

√(27.9/5.21) = w = 2.31 rad/s

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3 years ago

These clouds are low in the sky and tend to cover most of the sky. Which weather conditions are most likely to occur when you se

arsen [322]

Answer:

Light rain, sleet, or snow

Explanation:

pls brainlyest

4 0

3 years ago

Read 2 more answers

Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a

Vsevolod [243]

Answer:

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by $d sin\Theta =1.22\lambda$ --------eqn 1

The angle through which the first minima is diffracted is given by $tan\Theta =\frac{y_1}{D}$ ---------eqn 2

As $\Theta$ is very small so we can write $sin\Theta =tan\Theta$

So from eqn 1 and eqn 2 we can write

$y_1=\frac{1.22\lambda D}{d}$ --------eqn 3

Here $y_1$ is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so $y_1=\frac{14.7}{2}=7.35 \ cm$

Now putting all these value in eqn 3

$d=\frac{1.22\lambda D}{y_1}$

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

8 0

3 years ago

This is for science

andriy [413]

Answer:

Look at this picture hopefully it helps

6 0

3 years ago

An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum

Tju [1.3M]

Answer:

The pressure is $P_2 = 4.25 \ a.t.m$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 11.2\ a.t.m$

The temperature is $T_1 = 299 \ K$

Let the first volume be $V_1$ Then the final volume will be $2 V_1$

Generally for a diatomic gas

$P_1 V_1 ^r = P_2 V_2 ^r$

Here r is the radius of the molecules which is mathematically represented as

$r = \frac{C_p}{C_v}$

Where $C_p \ and\ C_v$ are the molar specific heat of a gas at constant pressure and the molar specific heat of a gas at constant volume with values

$C_p=7 \ and\ C_v=5$

=> $r = \frac{7}{5}$

=> $11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )$

=> $P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2$

=> $P_2 = 4.25 \ a.t.m$

8 0

3 years ago