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3 years ago

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a spring with an unknown force constant is compressed as shown. If the original length of the spring is 60cm, determine (a)the s

pring force constant (b) the compressive stress in the spring (c) the strain of the spring

1 answer:

aalyn [17]3 years ago

6 0

<span>(b) the compressive stress in the spring </span>

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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 is the peacocks displacement from 2 to 3 seconds

mash [69]

1.5 second difference

7 0

2 years ago

A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame

Dima020 [189]

Explanation:

It is given that,

length of steel wire, l = 0.75 m

Mass of the wire, m = 12 g = 0.012 kg

Fundamental frequency, f = 120 Hz

We need to find the mass of the anvil (m'). The fundamental frequency is given by :

$f=\dfrac{v}{2l}$

v is the speed of the mass

Speed is given by :

$v=\sqrt{\dfrac{T}{\mu}}$

$\mu$ is the mass per unit length, $\mu=\dfrac{m}{l}$

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

T is the tension in the wire,

$f=\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

$T=4f^2lm$

$T=4(120)^2\times 0.75\times 0.012$

T = 518.4 N

Tension in the wire, T = m' g

$m'=\dfrac{T}{g}$

$m'=\dfrac{518.4}{9.8}$

m' = 52.89 kg

So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.

6 0

3 years ago

Explain in your own words how the Doppler Effect is also applicable in our study of light.

arlik [135]

well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.

so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.

there hoped this helped I guess

8 0

3 years ago

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second

shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is 27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC = 25cos 30

OE = 25sin30

OE = CD

Therefore BC = 30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in $\Delta BCO$

$OB^2 = BC^2 + OC^2$

PUTTING ALL VALUE IN ABOVE EQUATION

$d^2 = 930-25sin30)^2 + (25cos30)^2$

$d^2 = 775$

d = 27.83 Km

distance between both oasis ( 1 and 2) is 27.83 Km

3 0

3 years ago