3 years ago

A speaker vibrates at a frequency of 200 Hz. What is its period

Physics

1 answer:

Daniel [21]3 years ago

8 0

Period, T = 1/ f.
f = frequency = 200 Hz.

Period T = 1/200 = 0.005 seconds.

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If an object moves twice as fast how do you think its energy changes? halves

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Answer:

If an object moves twice as fast its kinetic energy quadruples.

Explanation:

The kinetic energy (K₁) of a body of mass (m) that moves with speed (v) is:

K₁= 1/2 * m* v²

If we double the speed of the body, its kinetic energy (K₂) will be:

K₂= 1/2 * m*( 2v)²

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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

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3 years ago