Carbon capture is the safest way to capture emission.
<u>Explanation:</u>
Emission of green house gases is one of the most lurking threat above humans. So one of the best ways to reduce emission is to capture the gases like carbon di-oxide or carbon monoxide, nitrous oxide etc. So mostly, the carbon di-oxide gas which is produced as waste is captured using adsorption, absorption, chemical looping etc.
In this method, they are captured from high source points like cement factory etc and then stored in a secluded region. Thus, the adsorption of these gases and then storing them in secluded region is the best option for capturing emission.
Answer:
Particles of matter are packed more tightly in the ground than in the air
Explanation:
the reason is that sound waves are vibrations of the molecules of the medium: therefore, if the particles of the medium are closer together (as in solids), the vibrations can be transmitted faster, and the wave can travel faster
Answer:
Total Energy consumed = 146.94 kwh
Total cost = rs 587.76
Explanation:
Given:
3 bulbs each 60 watt [for 3 hours daily]
4 fans each 100 watt [for 8 hours daily
1 electric heater of 2 kWh (2,000 watt) [for half hour daily
]
Rate = rs 4/Kwh
Find:
Total Energy consumed
Total cost
Computation:
Energy used = Power × time
3 bulbs each 60 watt [for 3 hours daily] = 3 x 60 x 3 x 31 = 16,740 watt = 16.74 kw
4 fans each 100 watt [for 8 hours daily = 4 x 100 x 8 x 31 = 99,200 watt = 99.2 kw
1 electric heater of 2 kWh [for half hour daily
] = 1 x 2000 x (1/2) x 31 = 31,000 watt = 31 kw
Total Energy consumed = 16.74 + 99.2 + 31
Total Energy consumed = 146.94 kwh
Total cost = 146.94 x 4
Total cost = rs 587.76
Answer:
(a) 1.054 m/s²
(b) 1.404 m/s²
Explanation:
0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ)) - μk·m·g·(1 - sin(θ)) = m·a
Which gives;
0.5·g·cos(θ) - μ·g·(1 - sin(θ) = a
Where:
m = Mass of the of the block
μ = Coefficient of friction
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration of the block
θ = Angle of elevation of the block = 20°
Therefore;
0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°) - μk×9.81×(1 - sin(20°) = a
(a) When the static friction μs = 0.610 and the dynamic friction μk = 0.500, we have;
0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°) - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²
(b) When the static friction μs = 0.400 and the dynamic friction μk = 0.300, we have;
0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°) - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².
Answer:
Explanation:
Using the diagram (see attachment) we extract the following position vectors:
Next step is to find unit vectors 
as follows:
Using the diagram we find the corresponding vectors Forces:
Equation of Equilibrium:
Comparing i , j and k components as follows:
Solving Above equation simultaneously we get:
