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2 years ago

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Ocean waves are observed to travel to the right along the water surface during a developing storm. A Coast Guard weather station

observes that there is a vertical distance from high point to low point of 4.6 meters and a horizontal distance of 8.6 meters between adjacent crests. The waves splash into the station once every 6.2 seconds. Determine the amplitude, wavelength, frequency and period of these waves. Write out the equation that governs this behavior.

1 answer:

Nuetrik [128]2 years ago

6 0

Answer:

The amplitude is 2.3 m

The Wavelength is 8.6 m

The frequency is 0.16 Hz

The time period is 6.25 sec

The equation that governs the behavior is $Y=(2.3)sin[(\frac{2\pi}{8.6} )x -(\frac{2\pi}{6.2} )t]$

Explanation:

The explanation is shown on the first uploaded image

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A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, $\lambda_1=8\times 10^{-6}\ C$

linear charge density of the other charge distribution, $\lambda_2=-6\times 10^{-6}\ C$

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.r.\epsilon_0}$

where:

$\lambda=$ linear charge density

r = radial distance from the center of wire

$\epsilon_0=$ permittivity of free space

Therefore,

$E_1=E_2$

$\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}$

$\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}$

$\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}$

$x=0.0629\ m$

∴at y=6.29 cm the charge of the two distribution will be equal.

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1 year ago

A 5.00-kg ball is hanging from a long but very light flexible wire when it is struck by a 1.50-kg stone traveling horizontally t

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consider the right direction as positive and left direction as negative.

M = mass of the ball = 5 kg

m = mass of stone = 1.50 kg

$V_{bi}$ = initial velocity of the ball before collision = 0 m/s

$V_{si}$ = initial velocity of the stone before collision = 12 m/s

$V_{bf}$ = final velocity of the ball after collision = ?

$V_{sf}$ = final velocity of the stone after collision = - 8.50 m/s

using conservation of momentum

M $V_{bi}$ + m $V_{si}$ = M $V_{bf}$ + m $V_{sf}$

(5) (0) + (1.5) (12) = 5 $V_{bf}$ + (1.50) (- 8.50)

$V_{bf}$ = 6.15 m/s

h = height gained by the ball

using conservation of energy

Potential energy gained by ball at Top = kinetic energy at the bottom

Mgh = (0.5) M $V_{bf}^{2}$

(9.8) h = (0.5) (6.15)²

h = 1.93 m

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1 year ago

Read 2 more answers

A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h

Misha Larkins [42]

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

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1 year ago

Drag each tile to the correct location.

In-s [12.5K]

Answer:

Look at the image please

Explanation:

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1 year ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

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1 year ago