What process makes this pattern on the ocean floor?

What Process Makes This Pattern On The Ocean Floor?

Answers

Answer 1
Plate tectonics


and the ocean floor
Bathymetry, the shape of the ocean floor, is largely a result of a process called plate tectonics.

please give me brainliest

Related Questions

define potential difference as used in electricity​

Answers

Answer:

Explanation:

Testing answer please do not delete

the diffrence in potential betwen two points that represents the work involved or the energy relesed in the transfe of a unit quantity of electicity from one point to the other

HELPPP


Which of the following statements about friction is FALSE .
1 ) Friction involves objects that are in physical contact .
2) Friction is less if surfaces are very smooth .
3 ) Friction is always in the direction of the motion .
4) Friction is caused by the uneven surfaces of touching objects

Answers

Friction is caused by the uneven surfaces of touching objects

Ex 3) A tennis player lobs the tennis ball up in the air during a serve at a rate 3 m/s.
How long will it take the tennis ball to reach its maximum height?

Answers

10 seconds is the time it will take to reach maximum height

Computes average speed when a person walk for 1 minute at speed of 1.5 m/s and another 1 minute at speed of 3.5 m/s along a straight road

Answers

Answer:

2.5 m/s

Explanation:

(60*1.5+60*3.5)/120=2.5

or

(1.5+3.5)/2=2.5

ALREADY SOLVED Thank you for looking.

Answers

what nonsense you are say it have been solved when why did you put

Answer:

yes

Explanation:

A massive launcher sends a projectile vertically upwards from the surface of a planet of mass M and radius R. You may assume that this planet has no atmosphere. Part A If the projectile is launched at the escape speed, how do the magnitudes of its initial KE and gravitational potential energy (GPE) compare

Answers

Answer:

It can be shown that the potential energy of an object at the surface of the planet would be -G M / R if the potential at infinity is chosen to be zero.

Kinetic energy of G M / R would be required for the escape speed of such an object. The total energy in all such cases is zero.

This can easily be seen by considering the speed of an object falling from infinity towards the planet - the total energy will remain zero if it was zero when the object started to fall.

The potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.

What is escape speed?

Escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.

It is proven that if the potential at infinity is set to zero, the potential energy of an item on the planet's surface is [tex]\rm \frac{-GM}{R}[/tex].

The escape speed of such an item would necessitate kinetic energy will be [tex]\rm \frac{GM}{R}[/tex]. In all of these circumstances, the total energy is zero.

Consider the speed of an item falling from infinity towards the planet: if the total energy was zero before the thing began to descend, the total energy will stay zero.

Hence the potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.

To learn more about the escape speed refer to the link;

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Ex 3) A tennis player lobs the tennis ball up in the air during a serve at a rate 3 m/s.
How long will it take the tennis ball to reach its maximum height?

Answers

Explanation:

s=(v^2-u^2)/2g

=(0-3^2)/2*10

= 0.45m

4. What is the density of a block with a mass of 36 g and a volume of 9 cm?
O A 45 g/cm3
O B.27 g/cm
O 0.4 g/cm
O D. 0.25 g/cm

Answers

Answer:

0.4 g/cm

Explanation:

density (g cm ³) = mass (g)

÷

volume (cm³)

Answer:

C. 4 g/cm

Explanation:

Use the formula:

density = mass ÷ volume

mass = 36volume = 9

Sub in the values:

density = 36 ÷ 9 = 4 g/cm

Answer = 4 g/cm

A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary on a surface with a coefficient of kinetic friction of 0.15 when hit, how far does it move after the bullet emerges?

Answers

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

mass of the bullet, m = 23 g = 0.023 gspeed of the bullet, u = 230 m/smass of the wood, m = 2 kgfinal speed of the bullet, v = 170 m/scoefficient of friction, μ = 0.15

The final velocity of the wood after the bullet hits is calculated as follows;

[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s[/tex]

The acceleration of the wood is calculated as follows;

[tex]\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2[/tex]

The distance traveled by the wood after the bullet emerges is calculated as follows;

[tex]v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m[/tex]

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

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Someone help this is a test and i need to finish this quick i will give brainliest of it lets me

Answers

Sorry, I don't know but I think the correct answer is the first option.

A__________ shows how velocity changes over time

Answers

Velocity time graph goes into the blank hope it helps please mark me brainliest

Pls help need asap thank u!

Answers

Answer:

c

Explanation:

HELP! I need some assistance

Answers

Answer:

I'm a Filipino hahahah I'm not

10. Unless a light ray comes into contact with a surface or enters a different material, it
travels in a

Answers

Answer:

Straight Path

1) A bowling has a mass of 5 kg and a speed of 8 m/s. What is its momentum?
2) If the bowling ball started from a momentum of O and reached a momentum of 40
kg*m/s in 2 seconds. What is the impulse impacted on the bowling ball?
3) What is the force of impact for the 2 seconds?

Answers

Hgihucycyvyvuvuvuvuvuvuvuvvuvuvuv

A 3000 kg car stops at a red light, and is rear-ended by a 5000 kg truck traveling at 20m/s. In the collision, the two cars stick together. What is the final speed of the two cars just after the collision in m/s (Numeric Answers only)​

Answers

Explanation:

this is actually not as simple as it sounds here.

quite some energy is lost in the deformation of the bodies of car and truck, and it also needs more energy to get a standing object going than to accelerate an already moving object.

but assuming the simple described circumstances, then the energy and impulse of the moving truck of 5000 kg is transferred to a new combined system of car and truck of now 5000 + 3000 = 8000 kg.

so, the 20m/s inertia energy of the truck is now distributed to the truck/car combination.

since the same energy has to move now more mass, it is clear that the combined speed will be lower.

20×5000 = x×8000

20×5 = x×8

x = 100/8 = 12.5 m/s

that is the resulting speed of the combined truck/car object.

There are 6 foundation of sports and which one you think is the most important?

Answers

I just need the points

Explanation:

But just pick any and say something like "it stans out to me most/more" or "it sounds/looks more intristing to me"

Based on your observations, what can you say about your prediction in Parts A and B above concerning the potential and kinetic energy?

Answers

Answer:

Explanation:

My perdiction was correct. I predicted in Part A that the Kinetic Energy would increase. In Part B I predicted that the potential energy would decrease due to the kinetic energy increasing.

Un avión vuela horizontalmente con una velocidad de 800 km/h y deja caer
un proyectil desde una altura de 500 respecto al suelo. a) ¿Cuánto tiempo
transcurre antes de que el proyectil se impacte en el suelo?; b) ¿Qué distancia
horizontal recorre el proyectil después de iniciar su caída?

Answers

The projectiles launch allows to find the results for the questions of the launch of the bomb are:

     a) The fall time is: t = 10.1 s

     b) The distance traveled is: x = 2.24 10³ m

Projectile launching is an application of kinematics where the acceleration on the x-axis is zero and the acceleration on the y-axis is gravity acceleration.

a) Let's find the time until we reach the ground.

The initial vertical velocity is zero, the initial height is I = 500 m, when reaching the ground its height y = 0.

        [tex]y = y_o + v_o_y - \frac{1}{2} g t^2[/tex]  

        0 = y₀ + 0 - ½ g t²

        t = [tex]\sqrt{\frac{2y_o}{g}[/tex]  

        t = [tex]\sqrt{\frac{2 \ 500}{9.8} }[/tex]  

        t = 10.1 s

b) We look for the horizontal distance traveled.

Let's reduce to the international measurement system (SI).

       v = 800 km / h ([tex]\frac{1000m}{1km}[/tex]) ([tex]\frac{1h}{3600s}[/tex]) = 222.22 m / s

       x = vₓ t

       x = 22.22 10.1

       x = 2.24 10³ m

In conclusion, using the projectile launch relationships, we can find the results for the questions that are:

     a) The fall time is: t = 10.1 s

     b) The distance traveled is: x = 2.24 10³ m

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When converted to a household measurement, 9 kilograms is approximately equal to a

Answers

Ificicicucuivvicicicucucucu

Answer:

D) 19.8 lbs

Explanation:

1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.

1kg is also equal to 2.205 lbs. 9*2.205=19.8416

9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system

a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.

This leaves us with 19.8 lbs

10 basic rules of badminton?​

Answers

Explanation:

Rules

A match consists of the best of three games of 21 points.

The player/pair winning a rally adds a point to its score.

At 20-all, the player/pair which first gains a 2-point lead wins that game.

At 29-all, the side scoring the 30th point wins that game.

The player/pair winning a game serves first in the next game.

A badminton match can be played by two opposing players (singles) or four opposing players (doubles).

A competitive match must be played indoors utilising the official court dimensions.

A point is scored when the shuttlecock lands inside the opponent's court or if a returned shuttlecock hits the net or lands outside of the court the player will lose the point.

At the start of the rally, the server and receiver stand in diagonally opposite service courts.

A legal serve must be hit diagonally over the net and across the court.

A badminton serve must be hit underarm and below the server's waist height with the racquet shaft pointing downwards, the shuttlecock is not allowed to bounce. After a point is won, the players will move to the opposite serving stations for the next point.

The rules do not allow second serves.

During a point a player can return the shuttlecock from inside and outside of the court.

A player is not able to touch the net with any part of their body or racket.

A player must not deliberately distract their opponent.

A player is not able to hit the shuttlecock twice.

A 'let' may be called by the referee if an unforeseen or accidental issue arises.

A game must include two rest periods. These are a 90-second rest after the first game and a 5-minute rest after the second game.

The volume of an ideal gas is increased from 0.6 m3 to 2.4 m3 while maintaining a constant pressure of 1000 Pa (1000 N/m2). Determine, in J, the amount of work done by the gas in this expansion.

Answers

The amount of work done by the gas in the given expansion is 1800 J.

The given parameters;

initial volume of the ideal gas, V₁ = 0.6 m³final volume of the ideal gas, V₂ = 2.4 m³constant pressure of the gas, P = 1000 Pa

The amount of work done by the gas in the given expansion is calculated as follows;

W = PΔV

where;

ΔV is the change in volume of the gas

Substitute the given parameters and solve for the work done;

W = 1000(2.4  -  0.6

W = 1800 J

Thus, the amount of work done by the gas in the given expansion is 1800 J.

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PLEASE ANSWER ASAP IM GONNA MARK AS BRAINLIEST

What is the value of the force measured by the spring balance? ​

Answers

Spring balance is a mechanical device used for measuring the weight or force of an object by opposing the force of gravity with the force of an extended spring. The principle behind its working is the principle of Hooke's law.

Ex 1) A Major League baseball pitcher throws a baseball straight up into the air and the ball travels for 4
seconds reaching a height of 52 m before returning to the pitchers hand exactly at the height that he released
it. With what velocity does the baseball strike the pitchers hand on the way back down?

Answers

Answer:

52 / 4 = 13 m/s

Explanation:

Suppose a ball is dropped from a building. What is the change of velocity between the first and fifth seconds of its flight

Answers

Answer: 39.2m/s

Explanation:

Due to the free falling body, the acceleration of the ball is 9.8 m/s^2, and its initial velocity is 0 m/s.

V(t) = g*t

The change of velocity from t=1 to t=5:

Delta V = 9.8*(5-1) = 39.2 m/s

A spring of spring constant k = 200 N m−1 is slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. Calculate the work done by the extending force. 60

Answers

Answer:

31

Explanation:

No need

The correct answer is 31 I did it

What element is chemically similar to carbon

Answers

nonmental

that should be the answer

Car A uses tires for which the coefficient of static friction is 0.384 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 16.7 m/s. Car B uses tires for which the coefficient of static friction is 0.809 on the same curve. What is the maximum speed at which car B can negotiate the curve

Answers

Car C would like a summary too

A person is driving a car in a roundabout at 20 m/s so that its centripetal acceleration is 5 m/s2 . How much time it takes to complete 1 circle around the roundabout?

Answers

Answer:

Ac = v²/r

where, Ac is centripetal acceleration

v is velocity

r is radius

5 = (20)²/r

r = 400/5

r = 80m

from,

w = 2πf

where, w = angular velocity and f frequency

w = v/r

f = n/T

then,

v = 2πrn ,n = 1 to complete 1 cycle

T

T = 2πr/v

T = 2π*80/20

T = 25.14seconds.

The car would take 25.14s to complete 1 cycle of the roundabout.

*Sorry for the bad quality picture!*

A frictionless pendulum with a mass of 0.4 kg and a length of 2.1 m starts at point A, at an angle 0 of 60°. As it swings downward, it passes through point B, which is 30 degrees from equilibrium. What is the kinetic energy of the pendulum at point B?

A) 3.9 J

B) 3.0 J

C) 1.1 J

D) 4.1 J

Answers

The conservation of mechanical energy allows finding the result for the speed of the pendulum when it is at 30º is:

       

The speed is: 3.88 m / s

The conservation of mechanical energy is a theorem of greater importance in physics and ordinary life, it states that if there is no friction force the total mechanistic energy remains constant at all points.

Mechanical energy is the sum of kinetic energy plus all potential energies. In the attachment we see a diagram of the pendulum's movement at the two points of interest.

They indicate that the pendulum is released from an initial angle of θ₁ = 60º, let's find the mechanical energy at that point.

      Em₀ = U = m g h

Where the height is measured from the lowest point of the movement.

      h = L - L cos tea1 = L (1 cos tea1)

The second point of interest occurs for θ₂ = 30º.

At this point part of the energy is indica and part gravitational potential.  

     [tex]Em_f[/tex]  = K + U₂

      [tex]Em_f[/tex] = ½ m v² + m g h ’

There is no friction in the system, therefore mechanical energy is conserved.

       Em₀ = Em₀_f

       mg L (1 - cos θ₁) = ½ m v² + m g L (1 - cos θ₂)

       v² = 2g L (cos θ₂ - cos θ₁)

Let's calculate.

   

       v² = 2 9.8 2.1 (cos 30 - cos 60)

       v² = 41.16 0.366

       v = 3.88 m / s

In conclusion using the conservation of mechanical energy we can find the result for the speed of the pendulum when it is at 30º is:

       

The speed is: 3.88 m / s

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Answer:

3.0 J

Explanation:

Just took the test

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