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2 years ago

What is an essential characteristic of an object in equilibrium?

Physics

1 answer:

bulgar [2K]2 years ago

5 0

Answer:

2) zero acceleration

Explanation:

Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

Mathematically, the motion of an object is described in terms of time, distance, speed, velocity, position, displacement, acceleration, etc.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

Generally, an object is said to be in equilibrium when neither the energy possessed by the object not state of motion changes with respect to time. Thus, the vector sum of all the forces acting upon an object that's in equilibrium is zero.

In conclusion, an essential characteristic of an object in equilibrium is zero (0) acceleration because there's no change in its velocity with respect to time.

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What intitial action is the reason behind beginning a scientific investigation

Nataly [62]

To come to a conclusion would be the initial action that would be the reason behind beginning a scientific investigation.

When you want to prove your point and come to a conclusion, you would start an investigation. Your conclusion helps you form a ending statement.

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3 years ago

The sense of smell helps protect us against danger. Explain why.

atroni [7]

Well we can always smell things like if food smells bad or if we smell something we have to know whether that smell is good or bad. You could be able to smell animals that are dangerous. Or animal droppings, to know what animal it is if you are in danger.

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3 years ago

A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm . Both

kakasveta [241]

Answer:

The right response will be "450 volts".

Explanation:

The given values are:

R1 = 4.00 cm

R2 = 6.00 cm

q1 = +6.00 nC

q2 = −9.00 nC

As we know,

The potential difference between the two shell's difference will be:

⇒ $\Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]$

$=K[\frac{q1}{R2}-\frac{q1}{R1} ]$

On substituting the values, we get

$=(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]$

Δ $=450 \ volts$

3 0

3 years ago

____________ are used to calculate the distance a continent has moved in a year. a. Space satellites c. Laser beams b. Mirrors d

Law Incorporation [45]

<span> Space satellites, laser beams, mirrors</span> are used to calculate the distance a continent has moved in a year.

Therefore, your correct answer would be "all of the above".

4 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

2 years ago