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3 years ago

9

A lifeguard on a beach observes that waves have a speed of 7 m/s and a distance of 4.6m between wave crests. what is the period

of the wave motion?

1 answer:

Pepsi [2]3 years ago

4 0

Answer:

0.657 seconds

Explanation:

speed of wave= wavelength / time period

so

time period= wavelength / speed

= 4.6/7

=0.657 sec

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A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$

$V=\frac{2 m}{\rho} \\$

$V=\frac{2 \times 150}{1000} \\$

$\mathrm{~V}=0.3 \mathrm{~m}^3$

Area of pool $=\frac{\pi}{4} d^2$

$&=\frac{\pi}{4} \times 6^2 \\$

$&=28.27 \mathrm{~m}^2$

Height of the water rise

$h &=\frac{V}{A} \\$

$h &=\frac{0.3}{28.27} \\$

$& \mathrm{~h}=0.0106 \mathrm{~m}$

Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

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2 years ago

A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 12-m-long cable. You are (un

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3 years ago

Read 2 more answers

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

Korolek [52]

Answer:

a

$z= 2.5 \ m$

b

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

so

$z - (5-z) = 0$

=> $2 z - 5 = 0$

=> $z= 2.5 \ m$

Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

So

$z = \frac{5 + 3}{2}$

$z = 4\ m$

and

$z = \frac{ 5 -3 }{2}$

=> $z = 1 \ m$

=> $z = (1 \ m , 4 \ m )$

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