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Lina20 [59]
2 years ago
8

Name and Title:

Physics
1 answer:
Natalka [10]2 years ago
6 0

The variable that is altered by the independent variable is called the dependent variable. the spectrum given out by the hypothetical unknown celestial object.

Name : Christopher Andre

Instructor name : Albert brown

Name of the lab : ELS Laboratory

The term electromagnetic spectrum refers to the range of electromagnetic radiation's frequencies, as well as the wavelengths and photon energies connected to each frequency.

By analyzing the absorption spectra of the planets and moons, the electromagnetic spectrum experiment seeks to identify the components that make up their atmospheres.

From below one hertz to over 1025 hertz, electromagnetic waves are included in the electromagnetic spectrum.

The wavelengths that correlate to the frequency range from tens of thousands of kilometers to a small portion of the size of an atomic nucleus.

Starting at the low frequency (long wavelength) end of the spectrum, each frequency band's electromagnetic waves are referred to by a variety of names.

Hence the dependent variable is the one that changes as a result of the independent variable.

Learn more about electromagnetic spectrum here

brainly.com/question/13803241

#SPJ10

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A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At
STALIN [3.7K]

Answer:

(a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

Explanation:

Given that,

Nozzle diameter = 0.25 in = 0.00635 m

Hose pipe diameter = 0.64 in = 0.016256 m

Pressure = 1.9 atm =192518 Pa

(a). We need to calculate the speed of the water in the nozzle

Flow Speed at the inlet pipe will be given by using Continuity Equation

Q_{1}=Q_{2}

v_{1}A_{1}=v_{2}A_{2}

v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})

Where, A = area of pipe

A=\pi\times \dfrac{d^2}{4}

v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})

Put the value into the formula

v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}

v_{1}=3.014\ m/s

The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}

Where, h_{1}=h_{2}

P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2

P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)

Put the value into the formula

P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)

P_{1}=188081.702\ Pa

P=1.86\ atm

Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

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Answer:

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<em><u>Please mark as brainliest</u></em>

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Answer:

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Explanation:

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