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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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<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;  

Charge of electron e= 1.602 × 10^-19 C;  

kinetic energy E= 2.7 MeV

                          = 2.7 × 10^6 × 1.602 × 10^-19 J;

                          = 4.32 × 10^-13 Joules

But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

  = √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E =  4.32 × 10^-13 Joules

  r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

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3 0
3 years ago
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sp2606 [1]
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4 0
3 years ago
A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i
snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle=45^{\circ}

Acceleration due to gravity on earth is 9.8 m/s^2

Acceleration due to gravity on moon is \frac{9.8}{6}=1.63 m/s^2

Range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

R_{earth}=\frac{u^2\sin 2\theta }{g}=5----1

R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}-----2

Divide 1 & 2

\frac{5}{R_{moon}}=\frac{1}{6}

R_{moon}=30 m

4 0
3 years ago
The aeroplane lands at a speed of 80 m/s
strojnjashka [21]

The mass of the aeroplane is 300,000 kg.

<h3>What is Newton's second law of motion?</h3>

It states that the force F is directly proportional to the acceleration a of the body and its mass.

The law is represented as

F =ma

where acceleration a = velocity change v / time interval t

Given is the aeroplane lands at a speed of 80 m/s. After landing, the aeroplane takes 28 s to decelerate to a speed of 10 m/s. The mean resultant force on the aeroplane as it decelerates is 750 000 N.

The force expression will be

F = mv/t

Substitute the values and we have

750000 = m x  (80 -10)/ 28

750,000 = m x 2.5

m = 300,000 kg

Thus, the mass of the aeroplane is 300,000 kg.

Learn more about Newton's second law of motion.

brainly.com/question/13447525

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8 0
2 years ago
One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe
Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

\mu = \frac{m}{l}

\mu = \frac{8.79*10^{-3}}{70*10^{-2}}

\mu = 0.01255kg/m

The expression for the wavelength of the standing wave for the second overtone is

\lambda = \frac{2}{3} l

Replacing we have

\lambda = \frac{2}{3} (70*10^{-2})

\lambda = 0.466m

The frequency of the sound wave is

f_s = \frac{v}{\lambda_s}

f_s = \frac{344}{0.768}

f_s = 448Hz

Now the velocity of the wave would be

v = f_s \lambda

v = (448)(0.466)

v = 208.768m/s

The expression that relates the velocity of the wave, tension on the string and linear mass density is

v = \sqrt{\frac{T}{\mu}}

v^2 = \frac{T}{\mu}

T= \mu v^2

T = (0.01255kg/m)(208.768m/s)^2

T = 547N

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the n^{th} harmonic frequency is

f_n = nf_1

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

n=3

Then,

f_3 = 3f_1

Rearranging to find the fundamental frequency

f_1 = \frac{f_3}{3}

f_1 = \frac{448Hz}{3}

f_1 = 149.9Hz

7 0
3 years ago
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