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castortr0y [4]
3 years ago
5

One of the largest pipe organs is in the aŭditorium organ in the Fox Theatre. The pipe is 32

Physics
1 answer:
Evgen [1.6K]3 years ago
7 0

Answer:

Explanation:

Wave Period and Frequency In this video Paul Andersen explains how the period is the time between wave and the frequency is the number of ... Your browser does not currently recognize any of the video formats available. ... Show less Show more ... Waves: Light, Sound, and the nature of Reality.

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Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
3 years ago
The corpus callosum is comprised of more than 200 million axons that connect the hemispheres of the brain. Please select the bes
mamaluj [8]

Hello, love! The answer is True, or T, on Edge2020.

Hope this helped!

~ V.

8 0
3 years ago
Read 2 more answers
OSHA is the:
KATRIN_1 [288]
C Occupational Safety and Health Administration
8 0
2 years ago
How do we determine the conditions that existed in the very early universe? A We can only guess at the conditions, since we have
lakkis [162]

Answer:

D By looking all the way to the cosmological horizon, we can see the actual conditions that prevailed all the way back to the first instant of the Big Bang.

Explanation:

Astrophysicists are able to determine the conditions that existed in the early universe, by using instruments such as telescopes to observe and study cosmic horizons. More ideas about the early universe can be found from the thermal light present in cosmic backgrounds.

Scientists study these details that provide an insight into the conditions that existed so many years ago. They have been able to determine that the Big Bang involved so many collisions from these observations.

5 0
3 years ago
A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons
Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

5 0
3 years ago
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