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VladimirAG [237]
4 years ago
12

Figure 1 shows a wave movement during one second. What is the frequency of the wave

Physics
2 answers:
AURORKA [14]4 years ago
8 0
2 hertz, as explained previously- I'm afraid m/s is a completely different measurement
Andreyy894 years ago
4 0
This is 2 hertz.  You can mark out 2 full wavelengths in the second of time.
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The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r
Goshia [24]

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength \lambda =1.4cm =0.014m

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m

So distance D will be 49.34 m

4 0
3 years ago
If you slosh the water back and forth in a bathtub at the correct frequency, the water rises first at one end and then at the ot
notka56 [123]

Answer:

Velocity(v) = frequency(f) × wavelength

f = 0.3165

Wavelength = 2×length(L)

L = 157cm

Convert the length in centimetres to metre = 1.57m

v = 2×1.57 × 0.3165

v = 0.99m/s

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The velocity of a wave is the product of its frequency and it's wavelength. The frequency is already known. The wavelength is the distance between two successive wave crests which is formed by sloshing water back and forth in the bath tub. Sloshing water to one end of the tub will produce a wave crest first at that end then the other completing a cycle. The wavelength will be twice the length of the bath tub as it is the distance that both crests are formed.

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7 0
3 years ago
Helpp .....meeee pleaseee
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Answer:

hydrogen

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Explanation:

join this grop to get instant answer

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3 years ago
The efficiency of a device such as a lamp can be calculated using this equation:
loris [4]

efficiency = (useful energy transferred ÷ energy supplied) × 100

It's easy to use this formula, but we have to know both the useful energy and the energy supplied.  The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out.  I see two ways to do it:

<u>Way #1:</u>

We all know about the law of conservation of energy.  So we know that the total energy coming out must be  250J, because that's how much energy is going in.  The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.

(useful energy) / (energy supplied) =  (175J) / (250J) = <em>70% efficiency</em>

================================

<u>Way #2: </u>

How much of the energy is wasted ? . . . 75J wasted

What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted

30% of the input energy is wasted.  That leaves the other <em>70%</em> to be useful energy.

6 0
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SOVA2 [1]
If a mass of a neutron is 1 the electron mass is 0.00054386734 and it's charge is negative. Hope this helps! ;)
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