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2 years ago

PLEASE read this carefully.

1 answer:

5 0

For each of the given scenarios and based on the volume, mass, and density of the spheres, the water level is given as follows:

Scenario 1: The water level will rise; r

Scenario 2: The water level may rise or fall; r or f

Scenario 3: The water level will fall; f

<h3>What is the relationship between volume, density, and mass of the objects?</h3>

The density, volume and mass of an object are related by the formula below:

Density = mass/volume

The rise in the level of fluid when an object is placed in that fluid depends on the weight, volume and density of that object.

This is given by Archimedes' principle which states that the upthrust acting on a body immersed fully or partially in a fluid, is equal to the weight of the fluid displaced.

Based on the above, the water level in each of the scenarios is described thus:

Scenario 1: The new sphere has a radius of r < r0 and a mass of m > m0.

The new has a smaller volume and a greater mass and density. The water level will rise.

Scenario 2: The new sphere has a mass of m > m0 and a density of ρ < ρ0.

The new sphere has a greater volume and will float. Therefore, the water level may rise or fall.

Scenario 3: The new sphere has a density of ρ < ρ0 and a radius of r > r0.

The new sphere has a greater volume but less mass. Therefore, the water level will fall.

In conclusion, the rise or fall of the water level in the container depends on the volume, mass, and density of the spheres.

Learn more about Archimedes principle at: brainly.com/question/775316

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It’s measured in a reference frame that is usually the earth’s surface

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3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. <em>The correct form is: A </em> $+10\,\mu C$ <em> charge and a </em> $-10\,\mu C$ <em> at a distance of 0.3 meters. </em>

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$F$ - Electrostatic force, in newtons.

$\kappa$ - Electrostatic constant, in newton-square meters per square coulomb.

$|q_{A}|,|q_{B}|$ - Magnitudes of electric charges, in coulombs.

$r$ - Distance between charges, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C$ and $r = 0.3\,m$ , then the magnitude of the electrostatic force is: