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Annette [7]
3 years ago
7

What is the λ (in m) of an electromagnetic wave that has a frequency of 5.68 x 107 Hz? Enter a numerical response in 3 significa

nt figures; do not enter units.
Physics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

Wavelength of electromagnetic wave will be 5.281 m

Explanation:

We have given frequency of electromagnetic wave is f=5.68\times 10^7Hz

Velocity of electromagnetic wave is c=3\times 10^8m/sec

We have to fond the wavelength of the electromagnetic wave

We know that velocity of electromagnetic eave is given by v=\lambda f, here \lambda is wavelength and f is frequency

So \lambda =\frac{v}{f}=\frac{3\times 10^8}{5.68\times 10^7}=5.281m

Wavelength of electromagnetic wave will be 5.281 m

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Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much curre
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Answer:

4 A

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\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

Using the formula

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Hence, current flows through any one of the resistors is 4 A.

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3 years ago
Both Josef Loschmidt and Amedeo Avogadro contributed to our understanding of basic molecular numbers, sizes, and reaction ratios
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The  Avogadro’s number is used to represent the number of elementary entities that exist in one mole of a compound.

<h3>What is the  Avogadro’s number?</h3>

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2 years ago
Why do metals have similar properties?
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3 years ago
A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti
vovikov84 [41]

Answer:

<u>\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}</u>

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= 100 - 39.2

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10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}

10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}

10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}

\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4  

Frictional force on 40 kg slab by 10 kg block = 39.2 N  

40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}

40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}

40 kg slab will move with = 0.98 \mathrm{m} / \mathrm{s}^{2}

\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}

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