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3 years ago

What si the name of the state change water is going through as it boils?

Physics

1 answer:

Blababa [14]3 years ago

3 0

It evaporates into a vapor

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Which best describes a similarity between power plants that use water as an energy source and those that use wind as an energy s

disa [49]

The hydropower plant and wind turbines both uses kinetic energy to produce mechanical power and convert the mechanical energy using a generator to an electrical energy. They both have the process to produce energy but they differ in the source the hydropower plant uses water to whit the wind turbines power plant uses wind. Therefore the answer is letter B.

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3 years ago

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Eva's boyfriend says he loves her, but she wants proof. In order to obtain the most trustworthy nonverbal signals of how he real

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Answer:D) His Facial expressions when they spend time together

Explanation:

Eva's should carefully observe the normal verbal signs by her lover because facial expression gives the hint to listener or viewer that one is interested towards other.

Some common facial Expression are

Smiling Eyes:While smiling is, of course, an important expression when we speak, attempting to keep a constant smile could make you look more like a trained sociopath than a happy, enthusiastic presenter. These enables other person to feel the good vibes out of you and left a positive impact of your presence.
Nod on: It is important to express interest in the feedback of your audience. No one likes an identity-centered snob. One way to show that you're actually engaged in the comments and questions they pose is by nodding while you're listening. Moving your head a little wider will help to express your attention as well.

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3 years ago

A sound wave has frequency 620 Hz and wavelength 10.5 m. What is the speed of sound waves?

olasank [31]

Answer:

Explanation:

A sound wave in a steel rail has a frequency of 620 Hz and a wavelength of 10.5 m. What is the speed of sound in steel? U f = ? S f = 3 x 108 m/s) / 0.06 m 14 so it will be 14

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3 years ago

1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

love history [14]

Answer:

$ans \: = \boxed{{4.8 \times 10}^{ - 4} Hz}$

Explanation:

$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$

♨Rage♨

8 0

3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

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3 years ago