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3 years ago

15

If you were to double the separation between two slits, by what factor would the number of interference fringes within the centr

al maxima change?

1 answer:

NemiM [27]3 years ago

4 0

Answer The fringes become closer together as the slits are moved farther apart.

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Rainfall occurs on the earth due to effect of gravity

Anna [14]

<em>Answer:</em>

<h3><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>True</em></h3>

<em>Because </em><em>Gravity is the force of attraction between two objects, and Earth's gravity pulls matter downward, toward its center. It pulls precipitation down from clouds and pulls water downhill. Gravity also moves air and ocean water. ... Gravity pulls denser air and water downward, forcing less dense air and water to move upward.</em>

<em>Carryonlearning</em>

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2 years ago

If you had only one match, and entered a dark room containing an oil lamp, some newspaper, and some kindling wood, which would y

alukav5142 [94]

The match
you need to light the match before you can light anything else.

and after the match is lit, maybe light the oil lamp first

7 0

2 years ago

Read 2 more answers

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

2 years ago

A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of

Neko [114]

Answer: 185.5672566

Explanation: The friction is not relevant

Normal reaction is the force perpendicular to the surface.

this force resists the downwards forces applied which are gravity and a component of the applied force.

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2 years ago

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1 - It is okay to use slang on a record as long as the auditor can interpret it.

vesna_86 [32]

1. false 2. false 3. true 4. not sure 5. b 6. b or d 7. not sure 8.not sure 9. not sure 10. c

lol sorry if i’m wrong on any i’m just using common sense

6 0

2 years ago