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3 years ago

Rowenda is stopped on the side of the road. She then starts pedaling her bicycle for 15 seconds. Her final

Physics

1 answer:

lukranit [14]3 years ago

5 0

Explanation:

acceleration = change in velocity / change in time

a = v2-v1

- - - - - -

V1 = 0 ( since she stopped)

V2 = 9 m/s

t =15s

a = 9 - 0

- - - -

= 0.6m/s^2

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Two identical balls move directly toward each other with equal speeds. how will the balls move if they collide and stick togethe

Citrus2011 [14]

The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.

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3 years ago

Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.

pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is $D = 0.59 \ m$

Explanation:

From the question we are told that

The best resolution is $\theta = 0.3 \ arcsecond$

The wavelength is $\lambda = 700 \ nm = 700 *10^{-9 } \ m$

Generally the

1 arcminute = > 60 arcseconds

=> x arcminute => 0.3 arcsecond

$x = \frac{0.3}{60 }$

=> $x = 0.005 \ arcminutes$

Now

60 arcminutes => 1 degree

0.005 arcminutes = > z degrees

=> $z = \frac{0.005}{60 }$

=> $z = 8.333 *10^{-5} \ degree$

Converting to radian

$\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian$

Generally the resolution is mathematically represented as

$\theta = \frac{1.22 * \lambda }{ D}$

=> $D = \frac{1.22 * \lambda }{\theta }$

=> $D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }$

=> $D = 0.59 \ m$

4 0

2 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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3 years ago

Does gravity increase or decrease with greater distance?

Karo-lina-s [1.5K]

Decrease. Gravity is stronger here on earth than on the moon.

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3 years ago

Which of the following is true about acids?

Stells [14]

i would say, are sour and corrode metals

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3 years ago

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