E=274J
h=140cm=1,4m
g≈9,8m/s²

m=?

E=mgh
m=E/gh=274J/9,8m/s²*1,4m≈20kg

"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."

Regards M.Y.

3 0

3 years ago

Lightsail-2 is a spacecraft launched in June 2019 by the Planetary Society driven by radiation pressure on a solar sail of area

lozanna [386]

Answer:

$2.655\times 10^{13}\ photons$

Explanation:

Area of the solar sail = A = 30 m2

Solar constant = I = 1388 W/m2

Planck's constant = h = 6.626 × 10⁻³⁴ m²kg/s

Speed of light = c = 3×10⁸ m/s

Wavelength of light = $\lambda$ = 570 nm

Pressure from radiation

$P=2\frac{I}{c}\\\Rightarrow P=2\frac{1388}{3\times 10^8}\\\Rightarrow P=9.253\times 10^{-6}\ W$

Energy of a photon

$E=\frac{hc}{\lambda}\\\Rightarrow E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{570\times 10^{-9}}\\\Rightarrow E=3.487\times 10^{-19}\ J$

Number of photons

$n=\frac{P}{E}\\\Rightarrow n=\frac{9.253\times 10^{-6}}{3.487\times 10^{-19}}\\\Rightarrow n=2.655\times 10^{13}\ photons$

Number of photons is $2.655\times 10^{13}\ photons$

3 0

3 years ago

Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum da

FrozenT [24]

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light $\lambda =550nm=550\times 10^{-9}m$

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So $x=3\times 10^{-3}m$

We have to find the width of the slit

For the first order wavelength is equal to $\lambda =\frac{x}{D}\times a$ , here a width of slit

So $a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm$

So width of the slit will be equal to 1.47 mm

5 0

3 years ago