Answer:
(A) 140 j/sec (b) 1.26 K
Explanation:
We have given the heat heat flowing into the refrigerator = 40 J/sec
Work done = 40 W
(a) So the heat discharged from the refrigerator 
(b) Total heat absorbed =140 j/sec 
Let the temperature be 
Heat absorbed per hour =504000 ![[tex]=400\times 10^3\times \Delta T](https://tex.z-dn.net/?f=%5Btex%5D%3D400%5Ctimes%2010%5E3%5Ctimes%20%5CDelta%20T)
So 
Nitrogen oxides play a critical role in photochemical smog. They give the smog its yellowish-brown hue. Indoor residential appliances like gas stoves and gas or wood heaters can be significant emitters of nitrogen oxides in poorly ventilated environments.
- Nitrogen dioxide (NO₂), ozone (O₃), peroxyacetyl nitrate (PAN), and chemical compounds with the -CHO group are the main harmful elements of photochemical smog (aldehydes). If present in high enough amounts, PAN and aldehydes can harm plants and irritate the eyes.
- The greatest sources of emissions are power plants, heavy construction equipment driven by diesel, other moveable engines, and industrial boilers. Cars, trucks, and buses are next in line.
Therefore , on conclusion i.e. two gases with molecules consisting of nitrogen and oxygen atoms are nitric oxide (NO) and nitrogen dioxide (NO₂). These nitrogen oxides play a part in the development of smog and acid rain, adding to the issue of air pollution.
To know more about photochemical smog
brainly.com/question/15635778
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My guess would be about 10 years because stars are hot balls of light that are reflections from years ago so it would most likely take awhile
Answer
given,
For helium
Volume,V = 46 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₁ = ?
number of moles
we know
P V = n R T

n₁ = 1.89 moles
For oxygen
Volume,V = 12 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₂ = ?
number of moles
we know
P V = n R T

n₂ = 0.49 moles
Total volume of tank = 5 L
temperature of tank = 298 K
Partial pressure of helium


P₁ = 9.25 atm
Partial pressure of oxygen


P₂ = 2.39 atm
total pressure
P = P₁ + P₂
P = 9.25 + 2.39
P = 11.64 atm
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake