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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

#SPJ1

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(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find

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65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

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$f_{m_1}= 65 \ Hz$ , and

$f_{m_2}= 95 \ Hz$

Sampling rate $f_s = \ 245 \ Hz$

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s$

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz$

when n = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s$

$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$

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4 years ago

Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5

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Answer:

a) $432000\frac{N}{C}\hat{i}$

b) $-6.92\times10^{-14}N\hat{i}$

Explanation:

The magnitude of the electric field generated by a charged particle at a distance r is:

$E= k\frac{|Q|}{r^{2}}$

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is $25.0cm-5.0cm=20.0 cm=0.2m$ apart the positive charge is:

$E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}$

$E_1= 45000\frac{N}{C}$

and the electric field generated by q2:

$E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}$

$E_2=216000\frac{N}{C}$

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

$\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}$

$\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}$

Vector quantities satisfy superposition principle, this is $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$ , with E the total electric field.

$\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}$

b) The force is:

$\overrightarrow{F}=e*\overrightarrow{E}$ ,

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