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mr_godi [17]
3 years ago
7

What step of the rock cycle happens as the sea stacks are broken down

Physics
1 answer:
lesya692 [45]3 years ago
5 0

D. Weathering

The six steps of the rock cycle includes Weathering and Erosion, Transportation, Deposition, Compaction and Cementation. Metamorphism, Rock Melting.

Deposition is the process where rock particles sink at the rivers and become sediment.

Weathering is the process by which wind and water breaks down igneous, sedimentary, and metamorphic rocks. The process refers to large rocks broken down into smaller pieces and the broken off particles stay in the area.  


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a 0.500 kg mass is oscillating on a spring with k=330 N/m. the total energy of its oscillation is 3.24 J .what is the amplitude
juin [17]

Answer:

anyone know this or will i have to get my brother

6 0
3 years ago
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HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0
3 years ago
What happens to the coefficient of friction when the weight is increased? Why is this?
Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0
3 years ago
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A simple pendulum consists of a mass m (the "bob") hanging on the end of a thin string of length l and negligible mass. The bob
svetlana [45]

Answer:

The pendulum frequency  is (c) the same, or very close to it

Explanation:

The simple pendulum corresponds to a simple harmonic movement, to reach this approximation in the expression of the force the sine of the angle (θ) approaches an angle value, this is only true for small angles, generally less than 15º

Sine (15th) = 0.2588

The angle in radians is 15º π / 180º = 0.26180.2588 / 0.2618

The difference between these two values ​​is less than 1.2%

for smaller angle the difference is reduced more

Therefore, the period for both the 5º and 10º angles is almost the same

5 0
3 years ago
A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

8 0
3 years ago
Read 2 more answers
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