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Andru [333]
3 years ago
7

Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke

ck telescope?
Physics
1 answer:
nirvana33 [79]3 years ago
5 0

Answer:

4 times greater

Explanation:

<u>Step 1:</u> Calculate light-collecting area of a  20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² =\frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 20-m

A_{1} = \frac{\pi d^{2}}{4}

A_{1} = \frac{\pi (20^{2})}{4}

A₁ = 314.2 m²

<u>Step 2:</u> Calculate light-collecting area of a  10-meter Keck telescope (A₂)

A_{2} = \frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 10-m

A_{2} = \frac{\pi (10^{2})}{4}

A₂ = 78.55 m²

<u>Step 3</u>: divide A₁ by A₂  

= \frac{314.2 m^2}{78.55 m^2}

= 4

Therefor,  the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

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You have three identical metallic spheres, A, B and C, fixed to isolating pedestals. They all start off uncharged. You then char
kaheart [24]

Answer:

Charge on B is 12 uC.

Explanation:

Initial charge on A = 32 uC

Initial charge on B and C = 0

Now A touches to B, so the charge on A and B both is

q = (32 + 0) / 2 = 16 uC

Now A touches to C, so the charge on A and C both is

q' = (16 + 0) / 2 = 8 uC

Now again A touches to B so the charge on B is

q''= (8 + 16) / 2 = 12 uC  

7 0
3 years ago
T/F all compounds are molecules but NOT all molecules are compounds.
Solnce55 [7]
True because molecules don't have to be compounds but compounds have to be molecules
7 0
3 years ago
What is the period of a simple pendulum is its frequency is 20 Hz?
JulijaS [17]

Period  =  (1) / (frequency)

             =  (1) / (20/sec)

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5 0
4 years ago
You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric
marysya [2.9K]

Answer:

The Electric flux will be 0.42\times10^6\ \rm N.m^2/C

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

E=1.36\times10^6\ \rm N/C

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\

Let consider a  sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let \phi be the flux of the Electric Field coming out\passing through it which is given  by

\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C

It can be observed that same amount of  flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is  0.42\times10^6\ \rm N.m^2/C

8 0
3 years ago
Suppose you now take the ball and using a bat, pop it straight up into the air with a hang-time of 5.00 s (the hang time is how
dolphi86 [110]

Answer:

<h3>30.66m</h3>

Explanation:

Using the equation of motion formula S = ut + \frac{1}{2}gt^2 where;

S is the height to which the ball rises

u is the initial velocity of the ball = 0m/s

a is the acceleration due to gravity = 9.81m/s²

t is the time taken by the ball in air = 5.0s

Note that the  time to rise to the peak is one-half the total hang-time = 5.0/2 = 2.5s

Substituting the given parameters into the formula above to get S:

S = ut + \frac{1}{2}gt^2\\\\S = 0(2.5)+ \frac{1}{2}(9.81)(2.5)^2\\\\S = 0+\frac{1}{2}(9.81)\times 6.25 \\\\S = \frac{61.3125}{2}\\ \\S = 30.65625m\\\\S \approx 30.66m

This means that the ball rises 30.66m before it reaches its peak.

8 0
3 years ago
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