Match the term to its description.

Physics

1 answer:

Mekhanik [1.2K]3 years ago

6 0

Answer:

Elliptical galaxy

A) Forms a perfect sphere or an ellipse and is flattened to some degree

Irregular galaxy

B) Does not have a distinct regular shape

Lens galaxy

C) Has a large flattened core

Spiral galaxy

D) Has a central core from which curved arms spiral outward

The teems are already in the correct order

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With "normal" gravity, we used a potential energy of mgh. Now with the gravity that is more accurate over longer distances we us

Artemon [7]

Answer:

A general solution is $\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}$ and a particualr case is mgh, it is just to distance around the radius Earth.

Explanation:

We can use a general equation of the potential energy to understand the particular and general case:

The potential energy is defined as $U=-\int F\cdot dx$ , we know that the gravitational force is $F=GmM/r^{2}$ , so we could find the potential energy taking the integral of F.

$U=-GmM/r$ (1)

We can find the particular case, just finding the gravitational potential energy difference:

$\Delta U=U_{f}-U_{i}$ . Here Uf is the potential evaluated in r+Δh and Ui is the potential evaluated in r.

Using (1) we can calculate ΔU.

$\Delta U=-\frac{GmM}{r+\Delta h}+\frac{GmM}{r}$

Simplifying and combining terms we have a simplified expression.

$\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}$ (2)

Let's call $g=\frac{GM}{r^{2}}$ . It is the acceleration due to gravity on the Earth's surface, if r is the radius of Earth and M is the mass of the Earth and we can write (2) as ΔU=mgh, but if we have distance grader than r we should use (2), otherwise, we could get incorrect values of potential energy.