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3 years ago

The presently accepted value of the hubble constant gives an age of

Physics

1 answer:

koban [17]3 years ago

6 0

Answer:

14 billion years

Explanation:

The Hubble – Lemaître law, previously called the Hubble law, is a law of physics that states that the redshift of a galaxy is proportional to the distance it is, which is the same as, the further one galaxy is found from another, more quickly it seems to move away from it.

The Hubble constant is the value that measures the rate at which the expansion speed of the Universe varies with distance, and is one of the fundamental parameters of the Universe and allows, in particular, to determine the age of the Universe as we will see.

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Compare the geology of Callisto, Ganymede, and Titan.

V125BC [204]

Explanation: Ganymede, Callisto, Titan are the moons of outer planets or gaseous planets which are made up of ice and rock. Callisto is an ice-covered moon and has no inner or outer activity and is considered basically geologically dead. Ganymede has rocky core and shows signs of tectonic activity, including long cracks in the crust and regions of young surface terrain. Titan has active geology of liquid hydrocarbons on the surface, rain back onto the surface and evaporation into the atmosphere. It has similar size, composition and mass to Ganymede and Callisto.

8 0

3 years ago

10. Almost three-quarters of household energy use is due to the following:

Arturiano [62]

Answer: Energy consumption and sustainability is important so that it remain available for future generation.

Explanation:

1. The home furnaces are likely to require fuel like coal, which will directly emit carbon dioxide and carbon monoxide gases. These should be replaced with the electrical furnaces. The old or more power consuming air conditioners should be replaced with new ones.

2. The water heaters should be tankless so their capacity to heat more water could be possible. The water heaters should be electricity saving.

3. Washer and dryers should be water savy and electricity savy. A front-loading washing machine is useful energy saver.

4. The LED lights are more electricity saving than conventional bulbs. Halogen lights are also electricity saving.

7 0

3 years ago

Reflecting telescopes are popular because they're blank than a regarding telescope

olganol [36]

Reflecting telescopes are popular because they are more durable than a refracting telescopes.
There is a difference between these two types of telescopes, reflecting telescopes worked by reflecting light, they are also known as reflector and refracting telescopes worked by refracting light and they are also known as reflactor.
Reflecting telescopes also provides a good focus.

3 0

3 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

3 years ago

What is an example of velocity?What is an example of velocity?

givi [52]

Speed with direction

5 0

3 years ago

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