Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:
diameter = 40 mm ==> area = <em>π</em> (0.04 m)² ≈ 0.00503 m²
diameter = 120 mm ==> area = <em>π</em> (0.12 m)² ≈ 0.0452 m²
Then the speed at the end of the 40 mm pipe is
(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s
(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s
Answer:
The answer is 1250 mm
Explanation:
The path that particles move in the spectrometer is semicircular. Each of the particles has a displacement of twice the radius (2r) from the entrance to where the film is hit. According to the exercise, if the separation between the two molecules is 0.25 mm, then the difference in the radius of the molecules is equal to 0.125 mm. The ratio of mass/radius is equal for molecules, and therefore is equal to:
m = q * B * r / v
m/r = constant
(m/r)CO = (m/r)N = (28.0106 u/r) = (28.0134 u/(r + 0.000125 m))
Clearing and solving r:
r = 1.25 m = 1250 mm
Answer:
Sound waves are not transverse wave
Newton’s second law is that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.
The potential energy of the car when it let go is 20,000 J.
The speed of the car at the bottom of the ramp is 20 m/s.
The given parameters;
- <em>mass of the car, m = 100 kg</em>
- <em>height of the car, h = 20 m</em>
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The potential energy of the car is calculated as follows;
P.E = mgh
P.E = 100 x 10 x 20
P.E = 20,000 J
The speed of the car at the bottom of the ramp is calculated as follows;
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