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IrinaVladis [17]
3 years ago
12

A wave and a pendulum are both oscillators, why isn't a pendulum a wave?

Physics
1 answer:
Sladkaya [172]3 years ago
5 0
A pendulum cannot be considered a wave because it is attached to a point, a wave is an object that moves forward towards a certain point. 
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"A hiker starts at point P and walks 2.0 kilometers due east and then 1.4 kilometers due north. The vectors in the diagram below
Law Incorporation [45]
Here, you need to use your "Protractor" as it is given in the question, but we can calculate the value with the help of our mathematical calculation too:
[ Protractor can be use only in real life, not here ]

Draw an imaginary line from initial position to final position.
Now, In that triangle, tan x = P/B
tan x = 1.4 / 2
tan x = 0.70
x = tan⁻¹ (0.70)
x = 35  [ tan 35 = 0.70 ]

In short, Your Answer would be 35 degrees

Hope this helps!
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What is the best way to increase the solubility of something?
Svetllana [295]
INCREASE THE TEMPERATURE

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Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7
sleet_krkn [62]

Answer : \underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Explanation :

Given that,

Charge of particle 1 =  -1.50\times10^{-7} c

Distance x = 6 cm

Charge of particle 2 = 1.50\times10^{-7} c

Distance x = 27 cm

Total distance = \dfrac{6+27}{2}

r = 16.5\ cm

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm

Formula of electric field

E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}

Now, the the electric field due to  particle 1

\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i}  \dfrac{N}{C}

\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i}  \dfrac{N}{C}

\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Similarly, the electric field due to particle 2

\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Resultant Electric field

\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}

\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}

Hence, this is the required answer.






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The correct answer is B) Ultraviolet light source. Hope this helps.
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