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3 years ago

The image above shows a raging mountain river in full flood. What do you think is making the water brown in color?

Physics

1 answer:

MariettaO [177]3 years ago

4 0

Answer:

FloodPlains

Explanation:

The floodplains mix with the dirt in the water in which then makes it brown. You're welcome:)

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If frequency increases, then wavelength _________.

umka21 [38]

Answer:

Decrease

Explanation:

Please give me brainliest :)

8 0

3 years ago

In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

ANTONII [103]

Is there a picture?

6 0

3 years ago

A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai

Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air ( $\rho$ ), measured in kilograms per cubic meter, is defined by following equation:

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$M$ - Molar mass, measured in kilomoles per kilogram.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 2026.5\,kPa$ , $M = 28.965\,\frac{kg}{kmol}$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}$ and $T = 300\,K$ , then the density of air is:

$\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}$

$\rho = 23.534\,\frac{kg}{m^{3}}$

The mass of air stored in the vessel is derived from definition of density. That is:

$m = \rho \cdot V$ (2)

Where $m$ is the mass, measured in kilograms.

If we know that $\rho = 23.534\,\frac{kg}{m^{3}}$ and $V = 10\,m^{3}$ , then the mass of air stored in the vessel is:

$m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})$

$m = 235.34\,kg$

The mass of air stored in the vessel is 235.34 kilograms.

7 0

3 years ago

8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m fro

omeli [17]

The distance that the canoe moves in this process is 1.29 meters.

We first have to find the center of mass

$X = \frac{MsXs+McXc}{Mw+Mc} \\\\$

Where

Ms = Woman's mass = 45

Mc = Canoe's mass = 60kg

Xs = position from left= 1 cm

Xc = position from left end of canoe's mass = 2.5cm

When we put these values into the equation we have:

$X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\$

The center of gravity lies at the center of this boat. Therefore,

$Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5$

5.00 - 1. 00 = 4 meters

$\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\$

To get the distance that is moved by this canoe

distance = 3.143-1.857

= 1.286

≈ 1.29 meters

The distance that the canoe moves in this process is 1.29 meters.

6 0

2 years ago

Man A (70kg) and Man B (90kg) are hanging motionless from a platform at rest. What is the tension TA in the top rope if the plat

Damm [24]

Answer:

The tension in the upper rope (top rope), T1 = 1,888 N

Explanation:

The Parameters that were given:

Mass A, M1 = 70kg

Mass B. M2 = 90kg

acceleration, a = 2 m/s2

Assume the rope doesn't have mass, acceleration due to gravity, g

= 9.8 m/s2

The tension, T in a platform = m (a + g)

Then the tension, T1 in the upper rope = m1 (a + g) + T2

Where T2 = Tension in the lower rope

First, we calculate T2

Since the platform accelerates upward the acceleration would be positive

T2 = m2 (a + g)

T2 = 90kg ( 2 m/s2 + 9.8 m/s2)

T2 = 1,062N

To calculate the tension T1,

T1 = m1 (a + g) + T2

= 70kg (2 m/s2 + 9.8 m/s2) + 1062N

T1 = 1,888 N

3 0

3 years ago