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3 years ago

28. In 1.2 s, 52 wave crests spaced 0.12 m apart pass a location. What is the frequency of these waves?

Physics

2 answers:

Reptile [31]3 years ago

4 0

Answer:

0.7

Explanation:

Reasoning The distance between adjacent crests is the wavelength l. Since the speed of each wave is v = 3.00 × 108 m/s and the frequencies are known, the relation v = fl can be used to determine the wavelengths. The equation v = fl applies to any kind of periodic wave.

http://demo.webassign.net/ebooks/cj6demo/xlinks/c16-sec1-0002.htm

LenaWriter [7]3 years ago

3 0

Frequency = (number of waves) / (time)

(52waves) / (1.2 sec) = 43.333 Hz.

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3 years ago

This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ___________

melisa1 [442]

Answer:

have a component along the direction of motion that remains perpendicular to the direction of motion

Explanation:

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3 years ago

Determine the acceleration that result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object

user100 [1]

For 3 kg mass: F = $\lpha ma,a$ = $\frac{F}{m}$ = $\frac{12N}{3kg}$ = 4 $\lpha m$ / $s^{2}$

For 6 kg mass: F = $\lpha ma,a$ = $\frac{F}{m}$ = $\frac{12N}{6kg}$ = 2 $\lpha m$ / $s^{2}$

Hope this helped, have a great day!~

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3 years ago

A person kicks a ball off of a 50m high cliff with a speed of 10 m/s. How long will it take the ball to hit the ground? * 7 poin

Musya8 [376]

Presumably, the ball is kicked parallel to the ground below the cliff, so its altitude y at time t is

$y(t)=50\,\mathrm m-\dfrac12gt^2$

where g = 9.80 m/s^2 is the acceleration due to gravity.

The ball hits the ground when y = 0:

$0 = 50\,\mathrm m-\dfrac12gt^2$

$t^2=\dfrac{100\,\mathrm m}g$

$t=\dfrac{10}{9.80}\,\mathrm s\approx\boxed{3.2\,\mathrm s}$

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4 years ago

A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscill

Tasya [4]

Answer:

The amplitude of the oscillating electric field is 1316.96 N/C

Explanation:

Given;

frequency of the wave, f = 2.4 Hz

intensity of the wave, I = 2300 W/m²

Amplitude of oscillating magnetic field is given by;

$B_o = \sqrt{\frac{2\mu_o I}{c} }$

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

I is intensity of wave

c is speed of light = 3 x 10⁸ m/s

$B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T$

The amplitude of the oscillating electric field is given by;

E₀ = cB₀

E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶

E₀ = 1316.96 N/C

Therefore, the amplitude of the oscillating electric field is 1316.96 N/C

3 0

3 years ago