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3 years ago

Calculate the efficiency of an engine of the work input is 300 jewels and the work output is 100 rolls

Physics

2 answers:

monitta3 years ago

7 0

100/300=.33 33% efficient

irina1246 [14]3 years ago

6 0

Approximately 33% efficiency

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A star is moving toward the earth at 250 km/s. What can you say about the wavelength of the h-alpha absorption line that you wou

Juli2301 [7.4K]

The wavelength of the h-alpha absorption line that you would detect in the star's spectrum will be 54.67 * $10^{-2}$ nm

If the object is moving towards or away from us with some radial velocity, shifts can be observed in the location of the absorption

since, material is moving towards us a shift to shorter wavelength will be observed

shift = rest wavelength * ( radial velocity / speed of light )

= 656 * $10^{-9}$ * (250 * $10^{3}$ / 3 * $10^{8}$ )

= 54.67 * $10^{-2}$ nm

The wavelength of the h-alpha absorption line that you would detect in the star's spectrum will be 54.67 * $10^{-2}$ nm

learn more about radial velocity:

brainly.com/question/22409504?referrer=searchResults

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An object with a mass of 2.0 kg accelerates 2.0 m/s2 when an unknown force is applied to it. What is the amount of the force?

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so 2kg * 2m/s^2 = 4 N

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4 years ago

¿Que es un átomo???????

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3 years ago

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3 years ago

A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

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3 years ago