A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work)
1 answer:
Answer:
857.5 m
2.8583×10⁻⁶ seconds
Explanation:
Time taken by the sound of the thunder to reach the student = 2.5 s
Speed of sound in air is 343 m/s
Speed of light is 3×10⁸ m/s
Distance travelled by the sound = Time taken by the sound × Speed of sound in air
⇒Distance travelled by the sound = 2.5×343 = 857.5 m
⇒Distance travelled by the sound = 857.5 m
Time taken by light = Distance the light travelled / Speed of light
Time taken by light = 2.8583×10⁻⁶ seconds
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Answer:
W = 1.432 KJ
Explanation:
given,
mass = 22.2 Kg
angle of the rope = 27.5°
distance on the ground = 24 m
kinetic friction= μ = 0.32
acceleration due to gravity, g = 9.8 m/s²
Work done = ?
W = F d cosθ
a = 0 because it is moving with constant speed
equating all the forces acting in x direction
F cosθ = F friction = μN
equating all the forces acting in y direction
F sinθ + N -mg =0
now,
N = mg - F sinθ
putting value of N
F cosθ = μ mg -μ F sinθ
F (cosθ + μsinθ ) = μ mg
F =67.28 N
now,
W=F d cosθ
W =67.28 x 24 x cos(27.5)
W =1432.27 J
W = 1.432 KJ
Answer:
The net emissions rate of sulfur is 1861 lb/hr
Explanation:
Given that:
The power or the power plant = 750 MWe
Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:
= 3988126.8 J
= 3.99 MJ
Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.
i.e.
The mass of the coal that is burned per sec
The mass of the coal that is burned per sec = 187.97 lb/s
The mass of sulfur burned
= 2.067 lb/s
To hour; we have:
= 7444 lb/hr
However, If a scrubber with 75% removal efficiency is utilized,
Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr
= 0.25 × 7444 lb/hr
= 1861 lb/hr
Hence, the net emissions rate of sulfur is 1861 lb/hr