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3 years ago

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A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 29.4 s.

The distance between two successive crests is 31.4 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed

1 answer:

baherus [9]3 years ago

3 0

Answer:

(a) 5.88 s

(b) 0.17 Hz

(c) 31.4 m

(d) 5.338 m/s

Explanation:

From the question,

(a) Period = time between successive crest = t/n............ Equation 1

where t = time, n = number of wave crest

Given: t = 29.4, n = 5

therefore,

Period (T) = 29.4/5 = 5.88 s.

(b) Frequency = 1/period

Frequency = 1/5.88

Frequency = 0.17 Hz.

(c) wavelength = distance between two successive crest = 31.4 m

(d) using,

v = λf............... Equation 2

Where v = speed, f = frequency, λ = wavelength

given: f = 0.17 Hz, λ = 31.4 m

Substitute into equation 2

v = 0.17(31.4)

v = 5.338 m/s

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3 years ago

A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. How long does it take to come to rest? How many

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Answer:

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Explanation:

Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation

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3 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

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Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

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Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

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$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

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3 years ago