When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases. The opposite is also true—as frequency decreases, wavelength increases.
Answer:
P = 1000000[Pa] = 1000 [kPa]
Explanation:
To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.
where:
P = pressure [Pa] (units of pascals)
F = force = 100 [N]
A = area = 100 [mm²]
But first we must convert the units from square millimeters to square meters.
![A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2} } =0.0001[m^{2} ]](https://tex.z-dn.net/?f=A%3D100%5Bmm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7D%20m%5E%7B2%7D%20%7D%7B1000%5E%7B2%7Dmm%5E%7B2%7D%20%20%7D%20%3D0.0001%5Bm%5E%7B2%7D%20%5D)
Now replacing:
![P=100/0.0001\\P=1000000[Pa]](https://tex.z-dn.net/?f=P%3D100%2F0.0001%5C%5CP%3D1000000%5BPa%5D)
Thorns protect plants from insects or harmful bugs.
The position of the centre of gravity of an object affects its stability. The lower the centre of gravity (G) is, the more stable the object. The higher it is the more likely the object is to topple over if it is pushed. Racing cars have really low centres of gravity so that they can corner rapidly without turning over.
Increasing the area of the base will also increase the stability of an object, the bigger the area the more stable the object. Rugby players will stand with their feet well apart if they are standing and expect to be tackled.
(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then
(140 kg) (15 m/s) = (140 kg + 200 kg) <em>v</em>
==> <em>v</em> ≈ 6.2 m/s
(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:
<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>
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