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3 years ago

What is the frequency of this wave?

Physics

2 answers:

Musya8 [376]3 years ago

8 0

Answer:

its 2, not 4

Explanation:

brainlyest pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Reptile [31]3 years ago

3 0

It’s 4 because a coiled springs is closely spaced then widen

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Considering the factors that affect gravitational pull, in which location would the gravitational pull from the earth be SMALLES

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3 years ago

Describe the Compton Effect

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3 years ago

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Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

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$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

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3 years ago

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madreJ [45]

Explanation:

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2 years ago

What do electrons move from

Alona [7]

Answer:

Negatively charged, to positively charged parts

Explanation:

Electrons are negative, negative is attracted to positive.

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3 years ago