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Nat2105 [25]
3 years ago
7

The work of Subrahmanyan Chandrasekhar is likely to have been most influenced by the work of which of the following scientists?

A. Albert Einstein B. Edwin Hubble C. Marie Sklodowska-Curie D. Lord Kelvin
Physics
2 answers:
Alex73 [517]3 years ago
7 0

Edwin Hubble  ---apex

borishaifa [10]3 years ago
3 0

Answer

Correct Answer is B(Edwin Hubble)

Explantion

The work of Subrahmanyan Chandrasekhar is likely to have been most influenced by Edwin Hubble because both have worked in the feild of astrophysics.

Subrahmanyan Chandrasekhar was an astrophysicist. He belons to India. He is known for his work on the theoretical structure and evolution of stars. Subrahmanyan Chandrasekhar with William Fowler won the Nobel Prize in Physics in 1983 largely for this early work. He has worked in many other areas within theoretical physics and astrophysics.

Edwin Hubble was an astronomer who belongs to America. He is known for Hubble Law. Hubble discovered that many objects which was thought to be clouds of dust and gas  were actually galaxies beyond the Milky Way. He has an important contribution in establishing the fields of observational cosmology and extra galactic astronomy


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Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot
s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

f = \dfrac{nv}{2L}

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s  and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

f = \dfrac{1*343}{2*2.08}

f = \dfrac{343}{4.16}

f =82.45 \ Hz

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = \dfrac{nv}{2f}

The length of the longer pipe is L = \dfrac{1*343}{2*74.45}

The length of the longer pipe is L = \dfrac{343}{148.9}

The length of the longer pipe is L = 2.30 m

6 0
3 years ago
Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates
Alina [70]

Answer:

The capacitance of the deflecting plates is C=1.59 pF.

Explanation:

The expression for the capacitance of the capacitor in terms of area and distance is as follows;

C=\frac{\varepsilon _{0}A}{d}

Here, C is the capacitance, A is the area, d is the distance and \varepsilon _{0} is the absolute permittivity.

Convert the side of the square from cm to m.

s= 3.0 cm

s= 0.030 m

Calculate the area of the square.

A= s_^{2}

Put s= 0.030 m.

A=(0.030)_^{2}

A=9\times10^{-4}m^{-2}

Convert distance from mm to m.

d= 5.0 mm

d=5\times10^{-3}m

Calculate the capacitance of the deflecting plates.

C=\frac{\varepsilon _{0}A}{d}

Put d=5\times10^{-3}m, A=9\times10^{-4}m^{-2} and \varepsilon _{0}=8.85\times 10^{-12}Fm^{-1}.

C=\frac{(8.85\times 10^{-12})(9\times10^{-4})}{5\times10^{-3}}

C=1.59\times 10^{-12}F

C=1.59 pF

Therefore, the capacitance of the deflecting plates is C=1.59 pF.

4 0
3 years ago
_____________ and ____________ can fly in the stratosphere.<br><br> Fill in blanks
soldi70 [24.7K]
Jets and planes can fly in the stratophere


7 0
3 years ago
Read 2 more answers
A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-q
vovikov84 [41]

Answer:

Part a)

V = -1.52 V

Part b)

V = -1.16 V

Explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

V = \frac{kQ}{R}

here we know that

Q = Q_1 - 6Q_1

Q = - 5 Q_1

Q_1 = 2.70 pC

now we have

V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}

V = -1.52 V

Part b)

Potential on the axis of the ring is given as

V = \frac{kQ}{\sqrt{r^2 + R^2}}

V = \frac{(9\times 10^9)(-5\times 2.70\times 10^{-12})}{\sqrt{0.08^2 + 0.0671^2}}

V = -1.16 V

8 0
3 years ago
When an electron falls from a higher energy level to a lower energy level how is the energy released?
Illusion [34]
When a valence electron absorbs energy in the form of heat or light it uses that energy to jump to an excited state (outer level)
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3 years ago
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