All categories

3 years ago

Which of the following scientists won a noble prize of pioneering work in the study of the evolution of stars?

Physics

2 answers:

zhenek [66]3 years ago

7 0

Subrahmanyan chandrasekhar

Svet_ta [14]3 years ago

5 0

Subramanyan Chandrasekhar and William Alfred Fowler won a noble prize of pioneering work in the study of the evolution of stars.

The prize was divided equally between both scientists in 1983.

You might be interested in

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

3 years ago

The diagram shows two charged objects, X and Y.

masya89 [10]

The missing diagram is in the attachments.

Answer: X: positive Y: positive

Explanation: Electric field is a vector quantity, which means it can be represented by a vector arrow: the arrow points in the direction of electric field and its length represents the magnitude at a given location. There are another representation of the electric field called electric field lines, <u>in which the line points away from a positively charged source and towards a negatively charged source</u>. This occurs because it follows a pattern, where the lines points in the direction that a positive test charge would have if it is accelerating on the line.

Analyzing the diagram, it can be observed that the lines are pointing away from both of the charged objects. Therefore, both X and Y are <u>positively charged</u>.

4 0

3 years ago

Read 2 more answers

18. How much force would it take to push another, larger friend who has a mass of 70 Kg to accelerate

oksano4ka [1.4K]

Answer:

280N

Explanation:

f = mass x acceleration

4 0

2 years ago

The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate

butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * $10^{8}km$

speed of light = 3* $10^{8}m/s$

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 * $10^{8}*1000$ m

distance between earth and sun = 1.5 * $10^{11}$ m

speed = distance /time

time = distance / speed

time = $\frac{1.5*10^{11} }{3*10^{8} }$

$= 0.5*10^{3}$

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

3 0

3 years ago

A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w

slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0

3 years ago