What is a nebula? please help I am clueless

1 answer:

4 0

A nebula is gas, dust, and other things that form and form some sort of clould like thing. It also looks cool because of the colors it produces. I'm sure you could do a web search to find more about it, I don't take astronomy hahaha

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Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmosp

ankoles [38]

Answer:

Explanation:

Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.

P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

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3 years ago

What are the strengths and weaknesses of the four methods of waste management?<br>

Nitella [24]

Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;

* It creates employment

* It keeps the environment clean

* The practice is highly lucrative

* It saves the earth and conserves energy

The weaknesses of the methods of waste management includes;

* The sites are often dangerous

* The process is mostly

* There is a need for global buy-in

* The resultant product had a short life

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3 years ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,