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vekshin1
3 years ago
13

Kung hatiin sa tatlong pangkat ng 72 na bata isang bata mayroon sa bawat pangkat desisyon sentence​

Physics
1 answer:
podryga [215]3 years ago
8 0

Answer:

ewan ko papano bato pukinag ina

Explanation:

gagu ka kase putangina mo

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Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible
ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.    

Remember that in a circle the area is defined as:

A = \pi r^{2}  (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

     

d = 2r

r = \frac{d}{2} (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.  

   

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

A = \pi (\frac{d}{2})^{2}  (3)

Case for Large binocular telescope:

A_{mirror1} = \pi (\frac{8.4m}{2})^{2}    

A_{mirror1} = 55.41m        

For the second mirror will be the same value

A = A_{mirror1}+A_{mirror2}  

A = 55.41m+55.41m

A= 110.82m

Case for Keck 1 telescope:

A = \pi (\frac{10m}{2})^{2}    

A = 78.53m  

Case for Hobby-Ebberly telescope:

A = \pi (\frac{9.2m}{2})^{2}    

A = 66.47m  

Case for Subaru telescope:

A = \pi (\frac{8.3m}{2})^{2}    

A = 54.10m  

Case for Gemini North telescope:

A = \pi (\frac{8m}{2})^{2}    

A = 50.26m  

Case for Magellan 2 telescope:

A = \pi (\frac{6.5m}{2})^{2}    

A = 33.18m  

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

3 0
3 years ago
What are the four forms that water takes when it returns to earth from the clouds?
Marina CMI [18]
Precipitation from the sky to the earth can come in several forms. Four types are: 1) Rain 2) Snow 3) Sleet (think of a more liquidity slushy snow) 4) Hail (pretty much solid chunks of ice)
6 0
3 years ago
Please help me with this question​
Ksju [112]

Answer:

i can not read that sorry

4 0
3 years ago
How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?
Marrrta [24]

Answer:

9.2 amperes

Explanation:

Ohm's law states that the voltage V across a conductor of resistance R is given by V = R I

Here, voltage V is proportional to the current I.

For voltage, unit is volts (V)

For current, unit is amperes (A)

For resistance, unit is Ohms (Ω)

Put R = 12.5 and V = 115 in V=RI

115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes

8 0
4 years ago
A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin
Mashcka [7]

Answer:

r_{cm}=[12.73,12.73]cm

Explanation:

The general equation to calculate the center of mass is:

r_{cm}=1/M*\int\limits {r} \, dm

Any differential of mass can be calculated as:

dm = \lambda*a*d\theta  Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}

So, the differential of mass is:

dm = \frac{2M}{a\pi}*a*d\theta

dm = \frac{2M}{\pi}*d\theta

Now we proceed to calculate X and Y coordinates of the center of mass separately:

X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta

Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta

Solving both integrals, we get:

X_{cm}=2*a/\pi=12.73cm

Y_{cm}=2*a/\pi=12.73cm

Therefore, the position of the center of mass is:

r_{cm}=[12.73,12.73]cm

5 0
3 years ago
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