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2 years ago

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Um navio cargueiro proveniente do Oceano Atlântico passa a navegar nas águas menos densas do rio Amazonas. Em comparação com a s

ituação no mar, é correto afirmar que no rio: *

1 answer:

LUCKY_DIMON [66]2 years ago

8 0

Answer:

Mas eu acho q é o empuxo será igual e a porção imersa do navio será maior.

Explanation: SE

Professor marcos vê isso kkkkkkkkkkkkk

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A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

Read 2 more answers

If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice

Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

- Antares.

- Vega.

8 0

2 years ago

A windowpane is half a centimeter thick and has an area of 1.0 m2. The temperature difference between the inside and outside sur

polet [3.4K]

To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:

$\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)$

Where

k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)

$A = 1m^2$ Area

$L = 5*10^{-3}m$ Length

$T_H$ = Temperature of the "hot"reservoir

$T_C$ = Temperature of the "cold"reservoir

Replacing with our values we have that,

$\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)$

$\frac{Q}{t} = \frac{(0.84)(1)}{0.005} (15)$

$\frac{Q}{t} = 2520J/s$

Therefore the correct answer is B.

3 0

2 years ago

The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at

GenaCL600 [577]

Answer:

The speed of the light signal as viewed from the observer is c.

Explanation:

Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.

5 0

2 years ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used to solve this question.

From the diagram, the horizontal component of the force is:

$F_X = F_{cos \ \theta}$

Replacing 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

On the other hand, the vertical component is ;

$F_Y = Fsin \ \theta$

Replacing 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

$f = \mu \ N$

Where;

$\mu$ is said to be the coefficient of the friction

N = the normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing $F_Y \ for \ F_{sin \ \theta}$ . The normal reaction can now be:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Making $\mu$ the subject of the formular in the above equation; we have the following:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Replacing the following values: i.e

$F_X \ = \ 64.65 \ N$

m = 73 Kh

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Then:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Thus, the coefficient of friction is = 0.0984

5 0

2 years ago