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maw [93]
2 years ago
6

How much greater is the light-collecting area of a 6-meter telescope than a 3-meter telescope? How much greater is the light-col

lecting area of a 6- telescope than a 3- telescope?
A) two times
B) four times
C) six times
Physics
1 answer:
vovikov84 [41]2 years ago
7 0

Answer:

B) four times

Explanation:

Area of 6 meter telescope

A_1=\pi \frac{D^2}{4}\\\Rightarrow A_1=9\pi

Area of 3 meter telescope

A_2=\pi \frac{D^2}{4}\\\Rightarrow A_1=2.25\pi

Dividing the equations we get

\frac{A_1}{A_2}=\frac{9\pi}{2.25\pi}\\\Rightarrow A_1=4A_2

Hence, the 6-meter telescope has 4 times the light-collecting area of the 3-meter telescope.

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A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this
mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

7 0
3 years ago
A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what i
Neporo4naja [7]

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

N_p= 500 \ turn

N_s= 200 \ turn

I_s= 3.0 \ A

By using the transformer formula, we get

⇒ \frac{N_p}{N_s}  =\frac{I_s}{I_p}

⇒ I_p = I_s\times \frac{N_s}{N_p}

On substituting the given values, we get

⇒      =3.0 \ A\times \frac{2000}{500}

⇒      =12.0 \ A

8 0
3 years ago
Before Mt. Everest was discovered, what was the highest mountain in the world?
Juli2301 [7.4K]
Kangchenjunga (8,586 metres (28,169 ft)) was considered to be the highest mountain from 1838 until 1852. Mount Everest, 8,848 metres (29,029 ft). Established as highest in 1852 and officially confirmed in 1856.
3 0
3 years ago
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
lubasha [3.4K]

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

T = \frac{2\pi (r + h)}{v}

T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

M = 4.54 \times 10^{23} kg

Now the density is given as

\rho = \frac{M}{\frac{4}{3}\pi r^3}

\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

\rho = 12580.7 kg/m^3

8 0
2 years ago
Which statement describes the relationship between bond strength and the melting and boiling points of a substance? A. As the fo
icang [17]

Answer:

a

Explanation:

4 0
2 years ago
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