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1 year ago

When a wave is too steep to support itself, the wave front collapses creating a ________ that advances up the shoreline.

Physics

1 answer:

galben [10]1 year ago

8 0

When a wave is too steep to support itself, the wave front collapses therefore creating a break.

This is defined as the propagation of disturbance from one place to another in an organized manner.

In situations where the wave is too steep to support itself there is a break in the wavefront which advances up the shoreline.

Read more about Wave here brainly.com/question/6069116

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A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to

loris [4]

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of $V_{x}$ (horizontal component of velocity) and $V_{y}$ (vertical component of velocity)

We have

$V_{y}$ = Velocity in South direction = 18m/s

$m_{y}$ = Mass of piece in South direction = m

$V_{x}$ = Velocity in West direction = 18m/s

$m_{x}$ = Mass of piece in West direction = m

$V_{ry}$ = Velocity of Third Piece in North direction = unknown

$V_{rx}$ = Velocity of Third Piece in East direction = unknown

$m$ = Mass of third piece = 2m

$mV_{ry} = m_{y} V_{y} \\ 2mV_{ry} = m18\\ V_{ry} = 9$

$mV_{rx} = m_{x} V_{x} \\ 2mV_{rx} = m18\\ V_{rx} = 9$

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

$Hypotenuse^{2} = Height^{2} + Base^{2} \\ V_{3}^2 = 9^{2} + 9^{2} \\ V = 12.728 m/s$

V = 13m/s

6 0

3 years ago

Catalytic ozone destruction occurs in the stratosphere by a two-step reaction:

Sladkaya [172]

Catalytic ozone destruction occurs in the stratosphere where the reactions involving bromine, chlorine, hydrogen, nitrogen and oxygen gases form compounds that destroy the ozone layer. The reactions uses a catalyst (speeds up the reaction) in a two step reaction. considering chlorine the reactions appears as follows;
step 1
Cl + O3 = ClO + O2
step 2
ClO + O = Cl + O2
Where by chlorine is released to destroy the ozone layer, this takes place many times even with the other elements (hydrogen, bromine, nitrogen) and the end result is a completely destroyed Ozone layer

7 0

2 years ago

The piece of wood shown above has a mass of 20 grams. Calculate its volume and density. Then

kozerog [31]

We have that The Density and Volume is given as

$\rho=0.67\\\\\V=30cm^3$

The factors that affect Density are Mass and Volume

From the question we are told that

The piece of wood shown above has a mass of 20 grams.

Wood parameters

2cm

5cm

3cm

Generally the equation for the Volume is mathematically given as

$V=L*B*H\\\\\V=2*5*3\\\\V=30cm^3$

Generally the equation for the Density is mathematically given as

$\rho=\frac{m}{v}\\\\\\rho=\frac{20}{30}\\\\\rho=0.67$

Therefore the factors that affect Density are Mass and Volume

For more information on this visit

brainly.com/question/19694949?referrer=searchResults

4 0

2 years ago

In prokaryotic and eukaryotic cells the amount of water in the cell is kept at a steady state. The organelles

Len [333]

Answer:

Explanation:

It would not be C because prokaryotic cells don't have a nucleus. And if you're looking for a body in the cell that controls the nutrients and water it would be A.

3 0

2 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3

Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC Q₃ = 51 nC Q₂ = 30.5 nC

x₁ = - 1.715m x₃ = - 1.085m x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0

3 years ago