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Answer:
338 K
Explanation:
= 2.5 kg, c = 4189 J/(kg K), = 13.5 , = 22.5
= 0.5 kg, = 20
Heat loss by hotter water = heat gained by cooler water
cΔT = cΔT
c( - ) = c( - )
2.5 x 4189 x (22.5 - 13.5) = 0.5 x 4189 x ( - 20)
2.5 x 4189 x 9 = 2094.5 ( - 20)
94252.5 = 2094.5 - 41890
94252.5 + 41890 = 2094.5
136142.5 = 2094.5
=
= 65
= 65
But,
θ K = 273 + θ
= 273 + 65
= 338
The final temperature of the water is 338 K.
Climate feedbacks: processes that can either amplify or diminish the effects of climate forcings. A feedback that increases an initial warming is called a "positive feedback." A feedback that reduces an initial warming is a "negative feedback." So I guess it is ?
<span>Flow rate through pipe a is 0.4 m3/s
Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m
Length of Pipe a = 1000m
Length of Pipe b = 2650m
Temparature = 15 degrees
Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s
h = (f(LV^2)) / D2g
(fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb
LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2
Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2)
Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s
Vb = 3.4769 m/s
V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s</span>
Answer:
it will show a continuous rise in value. The rise will be sinusoidal.
Explanation: