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Allushta [10]
3 years ago
6

What are two types of diffraction?

Physics
1 answer:
tino4ka555 [31]3 years ago
7 0
Fresnel and Fraunhofer diffraction. Fresnel diffraction is produced when light from a point source meets an obstacle, the waves are spherical and the pattern observed is a fringed image of the object. Fraunhofer diffraction occurs with plane wave-fronts with the object effectively at infinity. The pattern is in a particular direction and is a fringed image of the source.
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What type of clouds usually accompany cold fronts?
ASHA 777 [7]

Answer: Cumulus

Explanation: Most large cloud fronts are made up of cumulus clouds, large storm clouds are cumulonimbus clouds.

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3 years ago
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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
vladimir1956 [14]

Answer:

Explanation:

kinetic energy required = 1.80 MeV

= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 2.88 x 10⁻¹³ J

If v be the velocity of proton

1/2 x mass of proton x v² = 2.88 x 10⁻¹³

= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

v² = 3.45 x 10¹⁴

v = 1.86 x 10⁷ m /s

If V be the potential difference required

V x e = kinetic energy . where e is charge on proton .

V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³

V = 1.8 x 10⁶ volt .

3 0
2 years ago
Explain how its possible for a compound to have both iconic and covalent bonds.<br><br>Thanks!
Evgesh-ka [11]
The atoms which make up the ion are covalently bonded to one another. 19) It is possible for a compound to possess both ionic and covalent bonding. a. If one of the ions is polyatomic then there will be covalent bonding within it.
3 0
3 years ago
Which statement about the physical properties of matter is true?
lisov135 [29]

I believe your answer would be D- Physical properties of matter effect the chemical properties of matter.

Hope this helps. Tell me if I'm right.

4 0
3 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
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