Having just enough weight to achieve all three states of buoyancy with only minor adjustments in the water is the definition of:

What is the precision (relative error) of the centripetal force divided by the mass if the velocity and the radius are each dete

Mrac [35]

The precision (relative error) of the centripetal force is 1%.

<h3>Relative error</h3>

This is the error in measurement of a variable obtained in comparison with other variables.

F = mv²/r

where;

F is centripetal force
m is mass
v is velocity
r is radius

F/m = v²/r

F/m = (0.01v)²/(0.01r)

F/m = 0.01v²/r

F/m = 1%(v²/r)

Thus, the precision (relative error) of the centripetal force is 1%.

Learn more about relative error here: brainly.com/question/13370015

5 0

3 years ago

A car with a mass of 1500 kg is pulled by a rope that is horizontal to the ground. The tension in the rope is 2000 N and a frict

Tanzania [10]

Answer:

Explanation:

Assuming the ground is level as well.

F = ma

a = F/m

a = (2000 - 350) / 1500

a = 1.1 m/s²

7 0

3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

3 years ago

The rising and setting of the Moon is mostly caused by *

frez [133]

Answer:

Hey mate......

Explanation:

This is ur answer.....

<h2><em>A. Rotation of Earth</em></h2>

<em>The moon rises in the east and sets in the west, each and every day. It has to. The rising and setting of all celestial objects is due to Earth's continuous daily spin beneath the sky</em><em>.</em>

Hope it helps!

Brainliest pls!

Follow me! :)

4 0

3 years ago

Read 2 more answers

Un'automobile di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

sesenic [268]

Answer:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

Explanation:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

3 0

3 years ago