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stiks02 [169]
3 years ago
8

Which does a reference point provide? Check all that apply.

Physics
2 answers:
Alona [7]3 years ago
6 0

A reference point is a point in geometry or general which is used to determine or refer to another point. In reference to this point, measurements can be done. Like, a reference point in space is used to measure distance or displacement to another point. A reference point in time is used to measure the time interval of an event. A reference in space-time can be used to measure the speed or detect object's motion. Hence, the following points apply:

→A position from which to measure future distance  

→a set of standard units for measuring displacement  

→a method for determining the speed of an object  

→a way to detect an object’s motion


A reference point provides a way/method to infer information about the object.


svetlana [45]3 years ago
6 0

Answer:

a position from which to measure future distance  

Explanation:

Reference point is a standard point taken as a fixed point for future measurement. Example, consider a ball is at height h. Here height h is measured from the surface of earth. So, the surface of earth is taken as a reference point for future measurement. In all measurements from that time is taken from the surface of the earth.

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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.

I hope i hleps you!

3 0
4 years ago
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Wittaler [7]

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Option"B" is correct.

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6 0
3 years ago
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NikAS [45]

Answer:

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Thus, option c is correct.

3 0
3 years ago
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