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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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1. Calculate the total heat content of 10 kg of ice at -23°C.?
vesna_86 [32]

To solve the problem, we must know the heat capacity of ice and water.

For Cp = 2090 J/kg C

H = mCpT

H = (10 kg) ( 2090 J/ Kg C) ( -23 C)

H = - 480700 J

For water Cp = 4180 j/kg C

H = (100 kg) ( 4180 J/kg C) ( 60 C)

<span>H = 2508000 J</span>

8 0
3 years ago
A person notices a mild shock if the current along a path through the thumb and index finger exceeds 82 µA. Find the maximum all
Drupady [299]

Answer:

V_{voltage}=19.68V

Explanation:

Given data

Current I=82µA=82×10⁻⁶A

Resistance R=2.4×10⁵Ω

to find

Voltage

Solution

From Ohms law we know that:

V_{voltage}=I_{current}*R_{resistance}\\V_{voltage}=82*10^{-6}*2.4*10^{5}   \\V_{voltage}=19.68V

4 0
3 years ago
A young kid of mass m = 36 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 3.5 m. The boy
lara31 [8.8K]

Answer:

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point mg-T=\frac{mv^2}{r}

where v=velocity needed to complete circular path

Th-resold velocity is given by mg-0=\frac{mv^2}{r}

v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s

So apparent weight of boy will be zero at top when it travels with a velocity of v=\sqrt{gr}

To get the velocity at bottom conserve energy at Top and bottom

At top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at Bottom E_b=\frac{mv_0^2}{2}

Comparing two as energy is conserved

v_0^2=4gl+gl

v_0^2=5gL

v_0=\sqrt{5gL}=13.09 m/s

Apparent weight at bottom is given by

W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N

6 0
2 years ago
When an electron is displaced in a semiconductor, the hole that's left behind is
morpeh [17]

Answer:

A. attracted to the negative terminal of the voltage source.

Explanation:

When an electron is displaced in a semiconductor, the hole that's left behind is

A. attracted to the negative terminal of the voltage source.

The electron leaving leaves a net + charge, which is attracted to the negative terminal.

8 0
2 years ago
2. Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 x 10-9 C and the oth
Maru [420]

Answer:

A. -2.16 * 10^(-5) N

B. 9 * 10^(-7) N

Explanation:

Parameters given:

Distance between their centres, r = 0.3 m

Charge in first sphere, Q1 = 12 * 10^(-9) C

Charge in second sphere, Q2 = -18 * 10^(-9) C

A. Electrostatic force exerted on one sphere by the other is:

F = (k * Q1 * Q2) / r²

F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²

F = -2.16 * 10^(-5) N

B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:

Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))

= - 6 * 10^(-9) C

Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C

Hence the electrostatic force between them is:

F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²

F = 9 * 10^(-7) N

7 0
2 years ago
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