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katen-ka-za [31]

3 years ago

9

Dissolved in many gasolines are cleaning agents and anti-knock additives. These cleaning agents and additives are an example of

A) solids. B) solutes. C) solutions. D) solvents.

2 answers:

Lesechka [4]3 years ago

7 0

I think the answer is Solutes.

Andrei [34K]3 years ago

5 0

Answer:

B

Explanation:

If ur doing usatestprep

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A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec

Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec$

Velocity of the string wave is equal to $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec$

Power of wave propagation is equal to $P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt$

So power of the wave will be equal to 5.464 watt

6 0

3 years ago

Put the waves in order from shortest to longest wavelength

FromTheMoon [43]

Answer:

b, a, c

Explanation:

The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.

3 0

2 years ago

Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. An

Vikentia [17]

Answer: Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Another bright star, Regulus, has a parallax of 0.042 arcseconds. Then, the distance in parsecs will be,23.46.

Explanation: To find the answer, we have to know more about the relation between the distance in parsecs and the parallax.

<h3>What is the relation between the distance in parsecs and the parallax?</h3>

Let's consider a star in the sky, is d parsec distance from the earth, and which has some parallax of P amount.
Then, the equation connecting parallax and the distance in parsec can be written as,

$d=\frac{1}{P}$

We can say that,

$dP=constant.\\thus,\\d_1P_1=d_2P_2$

<h3>How to solve the problem?</h3>

We have given that,

$d_1=2.6 parsecs.\\P_1=0.379arcseconds.\\P_2=0.042 arcseconds.\\d_2=?$

Thus, we can find the distance in parsecs as,

$d_2=\frac{d_1P_1}{P_2} =23.46 parsecs$

Thus, we can conclude that, the distance in parsecs will be, 23.46.

Learn more about the relation connecting distance in parsecs and the parallax here: brainly.com/question/28044776

#SPJ4

6 0

1 year ago

According to the Big Bang theory, when the universe first formed, the temperature was extremely hot. As it began to cool, proton

alex41 [277]

hydrogen would be the first Element

5 0

3 years ago

In 0.601 s, a 13.1-kg block is pulled through a distance of 4.19 m on a frictionless horizontal surface, starting from rest. The

Naya [18.7K]

Answer:

0.615 m

Explanation:

We need to determine the force on the spring first. By Newton's second law of motion, force is the product of the mass and acceleration. The mass is given.

The acceleration is determined using the equation of motion.

Given parameters:

Initial velocity, <em>u</em> = 0.00 m/s

Distance, <em>s</em> = 4.19 m

Time, <em>t</em> = 0.601 s

We use the equation

$s = ut+\frac{1}{2}at^2$

With <em>u</em> = 0.00 m/s,

$s = \frac{1}{2}at^2$

$a = \dfrac{2s}{t^2}$

$a = \dfrac{2\times4.19\text{ m}}{(0.601\text{ s})^2} = 23.2\text{ m/s}^2$

The force is

$F = (13.1\text{ kg})(23.2\text{ m/s}^2) = 303.92 \text{ N}$

From Hooke's law, the extension, <em>e</em>, of a string is given by

$e = \dfrac{F}{k}$

where <em>k</em> is the spring constant.

Hence,

$e = \dfrac{303.92\text{ N}}{494\text{ N/m}} = 0.615\text{ m}$

4 0

3 years ago