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What is the speed of a wave with a wavelength of 0.2m and a frequency of 1.5 Hz?
Wiki User
∙ 15y ago
The speed of sound of a wave c is wavelength lambda times frequency f. The speed of sound in that medium is c = 1500 meters per second. It could be water.
Wiki User
∙ 15y ago
Wiki User
∙ 10y agoAny wave in any medium, if its frequency is 1000 Hz and its wavelength
is 1.5 m, is propagating at 1,500 meters per second.
Wiki User
∙ 12y agoPeriod = 1/frequency = wavelength/speed = 1/20 = 0.05 second = 50 milliseconds
Wiki User
∙ 15y agov=f*lambda
v=0.2*1.5=0.3 m/s
Wiki User
∙ 15y agoC=λv C=(1.6 m) (400 1/sec) C= 640 m/sec
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What is he longest 20 cm 20 inches 200 mm 02m?
Convert them all into the same unit, for this case let's say cm. 20cm = 20cm 20 inches = 50.8cm 200mm = 20cm When you type 02m, do you mean 2m or 0.2m? 0.2m = 20cm 2m = 200cm If 0.2m, then 20 inches is the longest length. If 2m then that is the longest length.
How many times greater is 2 cm to 7 km?
7km is 350000 times greater than 2cm. Solution: 1. Clarify the question "How many times greater is 7km than 2cm?" 2. Convert both units into meters: 7*1000=7000m, 2*1/100=.02m 3. Divide the big number by the small number: 7000 / .02 = 350000 ====== Answer 1: 100 cm in a m, 1000 m in a km. 3.5 2 in 7. 3.5 * 100*1000 = 350000 centi-meters = 10**-2 meters kilo-meteres = 10**3 meters 2cm < 7km 2*10**-2/7*10**3 = (2/7)*10**-5
How much force is required to pump 1900 liters of water per minute through 20 mm nozzle to spray 50m in distance?
V= Q/A (velocity = volumetric flow rate / area) conversion: 1000L = 1m3; therefore 1900L/min= 1.9m3/min V = (1.9m3/min * 1min/60s) / Pi*(.02m)2/4 = 100.798 m/s Simplified Bernoulli Equation: ∆P/rho + 0.5*(initial velocity2-final velocity2) = 0 (rho=density) (equation assumes no fluid height changes and no frictional losses) ∆P/(1000 kg/m3) + 0.5*(02 - (100.798 m/s)2) = 0 ∆P = 5.08x106 N/m2= 5.08x106 Pascals P= F/A (Pressure= Force/Area) F= P*A F= ∆P * Area = 5.08x106 * Pi*(.02m)2/4 = 1595.97N A height of nozzle was not given; however, there is a minimum height when you specify a horizontal distance of 50m. Minimum nozzle height above ground to reach 50 meters horizontal distance: d=r*t (distance= rate * time) t= d / r t= 50m / 100.798m/s = 0.496s (time for a droplet to leave the nozzle and travel 50m horizontally) h= g*t2 (height of a falling object = gravity * time2 assuming spray nozzle is horizontal) h= 9.81m/s2 * (0.496s)2 = 2.41m above grade The minimum height will change if the nozzle is angled up or down. Note: 9.81m/s2 is approximately the earth's gravitational acceleration at sea level at the equator.
