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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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There is not enough information given to answer with. The force of gravity at the planet's surface depends on the planet's radius as well as its mass. The planet could have exactly the same mass as Earth has. But if it's radius is only 71% of Earth's radius, then gravity on its surface will be twice as strong as gravity on Earth.
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4 years ago
A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th
ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

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5 0
3 years ago
Please help ASAP!!
inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

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