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3 years ago

How does a longshore current change the beach

1 answer:

5 0

Longshore current brings about the transportation of sediments on the beach. Since waves usually approach the shore are an angle, the beach sand are carried up or down depending on the direction of the waves resulting in beach drift, that is, the net movement of the beach sand.

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A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered

Serjik [45]

Answer:

n = 1,732 the amplitude must be increased by a factor of 1,732

Explanation:

The power delivered by a wave is given by

P = E / t

P = ½ μ w² v A²

let's apply this expression to our case the power tripled

3P₀ = ½ μ w² v A’²

let's write the amplitude function of a initial amplitude

A ’= n A₀

where n is a number

3 P₀ = (½ μy w² v A₀²) n²

3P₀ = P₀ n²

n = √ 3

n = 1,732

therefore the amplitude must be increased by a factor of 1,732

4 0

3 years ago

An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

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<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

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<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

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<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

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<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0

3 years ago

Read 2 more answers

How many<br> different<br> elements are<br> present in<br> C6H12O6 ? How do you know

mestny [16]

3 elements
C6H12O6 contains 3 elements: carbon, hydrogen, and oxygen.

6 0

3 years ago

car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s

kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

$-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})$

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

$-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]$

4 0

3 years ago

The water that powers the generators enters and leaves the system at a low speed (thus we can neglect its change of kinetic ener

madam [21]

Answer:

Explanation:

Let density of water be ρ .

During flow , volume of water flowing per second is constant

loss of P. E per unit volume = ρ gh , 83.5 % is lost

Gain of K E per unit volume = 1/2 ρ v²

83.5 % of mgh = ρ 1/2 ρ v²

1/2 ρ v² = .835 x 9.8

v² = 2 x .835 x 9.8

= 16.366

v = 4.04 m /s

4 0

4 years ago