The mass of an electron is _____

Easy physics question help.!!!

stellarik [79]

Answer: This is not easy lol

Explanation:

3 0

2 years ago

Read 2 more answers

Longer wavelengths of light, such as _______, have ________ energy than shorter wavelengths, such as _________

Ilya [14]

Answer:

Micro and radio waves.

Lower energy.

Gamma rays.

Explanation:

The electromagnetic spectrum is the range of frequencies of electromagnetic radiation and their respective wavelengths.

Ionising radiation os defined as the energy required of photons of a wave to ionize atoms, causing chemical reactions.

The energy of the wave depends on both the amplitude and the frequency. If the energy of each wavelength is a discrete packet of energy, a high-frequency wave will deliver more of these packets per unit time than a low-frequency wave. In summary, the longer the wavelength, the lower the energy to ionise.

The velocity of a wave is directly proportional to the frequency of that wave.

c = f * lambda

Where,

c = velocity of the wave

f = frequency of the wave = 1/time

Lambda = wavelength.

From the above expression, the longer the wavelength, lambda the shorter the frequency.

Examples of waves with longer wavelengths are, micro and radio waves, while radiations with shorter wavelengths like gamma rays.

8 0

3 years ago

The distance between stars is typically measured in

Murljashka [212]

<span>The question says: complete the sentence: The distance between stars is typically measured in.... The answer is 'light years" - the distance that light travels in a year. That's because it's a very big unit and if we were using smaller units, we would be using huge numbers that would be hard to read and would take up a lot of space.</span>

3 0

3 years ago

Air pressure increases as you travel higher above sea level. This is the reason that cabins in commercial airliners require pres

irakobra [83]

The answer is true about the cabins in commercial airliners that require pressurization.

<h3>Why are the cabins of commercial airplanes pressurized?</h3>

Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.

So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.

Learn more about pressure here: brainly.com/question/28012687

#SPJ1

4 0

1 year ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(c) v = 0.31m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

3 years ago

The mass of an electron is ______