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barxatty [35]
3 years ago
9

"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro

ups dozens of millions of lightyears away. What method do astronomers use with the Hubble to find such distances
Physics
1 answer:
pickupchik [31]3 years ago
5 0

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

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A light-year is a unit of a. time. c. mass. b. distance. d. density. please select the best answer from the choices provided a b
11111nata11111 [884]

Answer:

distance

Explanation:

it is the distance traveled by light in one year

3 0
1 year ago
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For the following questions, imagine the book drops in free fall and falls a distance d. For the system of the book and earth, t
alukav5142 [94]

Answer:

C) is zero

Explanation:

According to the law of energy conservation, the total mechanical energy of the object is conserved. A book falling a distance d would have a change in potential energy, resulting in the same change in kinetic energy. But the total mechanical energy must be the same. So there's 0 change in total energy of the system.

8 0
3 years ago
JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back
zzz [600]
(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second 

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>
3 0
3 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
2 years ago
What is described by the terms body-centered cubic and face-centered cubic?
Montano1993 [528]
In physical chemistry, the terms body-centered cubic  (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.

It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.

7 0
3 years ago
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