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3 years ago

How do noise canceling headphones work

Physics

2 answers:

ivanzaharov [21]3 years ago

7 0

My guess would be that it interferes with sound waves.

LUCKY_DIMON [66]3 years ago

7 0

Answer:

there are different sound wave frequencies

Explanation:

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Number 25 and 26 please!!!

skad [1K]

laborious ... enclosed may help ... cos -angle = cos +plus angle ... sin - anfle = - sine angle

cos is even function, sine is odd ... draw rhe graohs

6 0

4 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago

100 grados a revoluciones

djyliett [7]

No. Quatro cientos grads.

6 0

3 years ago

The diagram shows a rectangular object acted on by three horizontal forces.

sammy [17]

<h3>right --- 20 + 10 = 30 N</h3>

<h3>resultant force --- 30 - 10 = 20 N</h3>

4 0

3 years ago

Read 2 more answers

slader) Two charges are arranged at corners of a square which has a side length of L = 0.25 m. The values of the charges are q1

nataly862011 [7]

Explanation:

Formula to calculate the electric potential is as follows.

$V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

Putting the given values into the above formula as follows.

$V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

= $\frac{9 \times 10^{9}}{0.25}[\frac{-3.3}{\sqrt{2}} + 4.2] \times 10^{-6}$

= $6.72 \times 10^{4} V$

Hence, electric potential at point A is $6.72 \times 10^{4} V$ .

Now, the electric potential at point B is as follows.

$V_{B} = \frac{kq_{1}}{L} + \frac{kq_{2}}{\sqrt{2}L}$

= $\frac{9 \times 10^{9}}{0.25} [-3.3 + \frac{4.2}{\sqrt{2}}] \times 10^{-6}$

= $-1.19 \times 10^{4} V$

Hence, electric potential at point B is $-1.19 \times 10^{4} V$ .

8 0

4 years ago