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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.

You might be interested in
Neglecting air resistance, what maximum height will be reached by an arrow launched straight upward with an initial speed of 35
tankabanditka [31]
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.

The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)

Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
3 0
3 years ago
Question 7 of 25
tiny-mole [99]

Answer:

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

2Na + MgCl₂ → 2NaCl + Mg

Explanation:

A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation

Among the given chemical reactions, we have;

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

In the above reaction;

The number of phosphorus, P, on either side of the equation = 2

The number of bromine atoms, Br, on either side of the equation = 6

The number of chlorine atoms, Cl, on either side of the equation = 6

Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced

The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg

In the above reaction, there are two sodium atoms, Na,  one magnesium atom and two chlorine atoms on both sides of the reaction, therefore, the reaction is balanced

6 0
2 years ago
A lava flow is an example of what igneous rock
nydimaria [60]
The answer is extrusive. 
7 0
3 years ago
Read 2 more answers
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
Artemon [7]

Answer:

15\%.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

4 0
2 years ago
A 3.5 kg object moving at 8.1 m/s in the positive direction of an x axis has a one-dimensional elastic collision with an object
Tamiku [17]

Answer:

So the mass of the second object M will be 1.951 kg    

Explanation:

We have given mass of the first object m_1=3.5kg and its velocity v_1=8.1m/sec

Mass of the second object m_2=M  it is at rest so its velocity v_2=0

From conservation of momentum we know that

Initial momentum = final momentum

So m_1v_1+m_2v_2=(m_1+m_2)v

3.5\times 8.1+M\times 0=(3.5+M)5.2

28.35=18.2+5.2M

M = 1.951 kg

8 0
3 years ago
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