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irina1246 [14]
4 years ago
14

11. What causes potential energy that accumulates in a fault to transfer to the surface during an

Physics
2 answers:
Llana [10]4 years ago
6 0

Answer: Seismic waves

gulaghasi [49]4 years ago
3 0

Answer: your answer is B

Explanation: Seismic waves causes potential energy that accumulates in a fault to transfer to the surface during an earthquake.

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A ball is kicked off of a roof at 23 m/s [R 25° U]. What is the height of
Ymorist [56]

Answer:

Explanation:

Considering the fact that we ave been given an angle of inclination here, we best use it! That means that the velocity of 23 m/s is actually NOT the velocity we need; I tell my students that it is a "blanket" velocity but is not accurate in either the x or the y dimension of parabolic motion. In order to find the actual velocity in the dimension in which we are working, which is the y-dimension, we use the formula:

v_{0y}=v_0sin\theta and filling in:

v_{0y}=23sin(25) which gives us an upwards velocity of 9.7 m/s. So here's what we have to work with in its entirety:

v_{0y}=9.7m/s

a = -9.8 m/s/s

t = 2.8 seconds

Δx = ?? m

The one-dimensional motion equation that utilizes all of these variables is

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 9.7(2.8)+\frac{1}{2}(-9.8)(2.8)^2 I am going to do the math according to the correct rules of significant digits, so to the left of the + sign and to 2 sig fig, we have

Δx = 27 + \frac{1}{2}(-9.8)(2.8)^2 and then to the right of the + sign and to 2 significant digits we have

Δx = 27 - 38 so

Δx = -11 meters. Now, we all know that distance is not a negative value, but what this negative number tells us is that the ball fell 11 meters BELOW the point from which it was kicked, which is the same thing as being kicked from a building that is 11 meters high.

6 0
3 years ago
Where would you weigh the most?
uranmaximum [27]

Answer:

mars

Explanation:

4 0
4 years ago
Read 2 more answers
Express 2759 mockingbirds in kilomockingbirds. answer in units of kmockingbird
Nadya [2.5K]
I would think 2.759kmockinbirds if you consider mockingbirds equal to a metre and kmockinbird to kilometres
6 0
3 years ago
An object is realsed From rest . how far does it fall during the second second of fall?​
Savatey [412]

Answer:

14.715 m

Explanation:

Assume that the acceleration due to gravity is 9.81 m/s^2 downwards, take downwards as positive

First second:

v = u + at

v = 9.81 m/s

Second second:

s = ut + (1/2)at^2

s = 9.81(1) + (1/2)(9.81)(1)^2

s = 14.715 m

6 0
3 years ago
point A is at the bottom of a rough plane which is inclined at angle tita to the horizontal . A body of mass M is projected from
ziro4ka [17]

<h2>The work done , when body moves along the plane </h2>

Explanation:

A body is projected from bottom of the inclined plane . When body is going up the plane .

The downward force = m g sinθ is developed due to its weight

As body is moving upwards , the force of friction will act downwards

The force of friction = μ R

here μ is the coefficient of friction ans R is the normal reaction

Thus force of friction f = μ mg cosθ

Let the acceleration upwards is a

The upward force required = m a

Thus m a = mg sinθ + μ mg cosθ

or acceleration a = g ( sinθ + μ cosθ )

The work done in moving upwards  W = F S

Thus W =  mg ( sinθ + μ cosθ ) S

here S is the displacement on the plane

When body moves down , the force of friction acts upwards

Thus m a = m g ( sinθ - μ cosθ )

The work done W = m g ( sinθ - μ cosθ ) S

As the body is projected with velocity u

which can be calculated by the relation v² - u² = - 2 a X

Here v = 0 at the highest point

Thus u = \sqrt{2ax}

here a = g ( sinθ + μ cosθ )

Similarly , when it moves down , the initial velocity u = 0

Thus v² - 0 = 2 a x

or  v = \sqrt{2ax}

here a = g ( sinθ - μ cosθ )

3 0
3 years ago
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