Answer:
α =
Explanation:
For this exercise we use Newton's equation for rotational motion
∑ τ = I α
the troque is
α = Fr .r
the moment of inertia of a cylinder is
I = ½ m r²
we substitute
fr r = (½ m r²) α
the expression friction is
fr = μ N
we substitute
μ N r = ½ m r² α
α =
Answer:
When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down
Answer:
2.47 %
Explanation:
We are given;
Frequency emitted by source(bird); f_s = 1420 Hz
Frequency heard by observer; f_o = 1456 Hz
From doppler shift frequency, we know that;
f_o = f_s [c/(c - c_s)]
Where c_s is speed of source which is the bird and c is speed of sound.
Thus;
Rearranging the equation, we have;
f_s/f_o = (c - c_s)/c
f_s/f_o = 1 - (c_s/c)
Plugging in the relevant values to get ;
1420/1456 = 1 - (c_s/c)
0.9753 = 1 - (c_s/c)
1 - 0.9753 = (c_s/c)
(c_s/c) = 0.0247
Since we want it expressed im percentage,
Thus, (c_s/c) % = 0.0247 x 100 = 2.47 %
<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>