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3 years ago

Describe a sea-level transgression

1 answer:

7 0

<span>Some geographic areas endure cycles between these two processes called transgressive-regressive sequences. The rocks of western Pennsylvania are one example. Sandy beaches often leave observable records of transgression by covering marsh sediments that were once behind it as it moves inland. The original sediments are then covered by even deeper water sediments, which geologists can trace and record. It is generally believed that transgression will increase in accordance with rising sea levels worldwide</span>

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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

Ira Lisetskai [31]

The elastic potential energy (Ep) is given by $Ep = \frac{1}{2}*k*x^2$

Data:
Ep = 5184 J
k = 16200 N/m
x (displacement) = ?

Solving:
$Ep = \frac{1}{2}*k*x^2$
$5184 = \frac{1}{2}*16200*x^2$
$5184*2 = 16200x^2$
$10368 = 16200x^2$
$16200x^2 = 10368$
$x^2 = \frac{10638}{16200}$
$x^2 = 0.64$
$x = \sqrt{0.64}$
$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\quad\checkmark$

8 0

4 years ago

Read 2 more answers

A field researcher uses the slow-motion feature on her phones camera to shoot a video of an eel spinning at its maximum rate. Th

posledela

Answer:

$\alpha=(1/120)*14*360=42$ degrees

Explanation:

Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way

So the angle the eel rotate is going to be

Camera record 120 frames per second

time taken for 1 frame record

$t_t= 1/120 seconds$

eel rotates

$\alpha_1=14rev *360$ degrees in 1 second

so:

eel rotetas $\alpha=(1/120)*14*360=42$ degrees

5 0

4 years ago

You were in a tog-o-war contest which your team won. Write a letter to your friend that tells why your team won the contest. Rem

ollegr [7]

Answer:

Imagine a tug-of-war between you and one

friend. If you are stronger, you apply more

force to the rope. You pull your friend across

the line, and you are the winner! If your friend

is stronger, he might pull you across the line.

Sometimes the forces are equal. Neither you

nor your friend moves across the line. The two

forces are balanced.

We say that the net force on an object is the combination of all the forces acting on it. To find the net force of

forces that are acting in the same direction, add them together. For example, if you pull on a box with a force of 25

newtons (N) while your friend pushes the box (in the same direction you are pulling) with a force of 30 N, the net

force applied to the box in that direction is 55 newtons.

To find the net force of forces that are acting in opposite directions, subtract the smaller force from the larger

one. If you are pulling on a tug-of-war rope with a force of 40 N, and your friend is pulling with a force of 35 N in

the opposite direction, the net force on the rope is 5 newtons in your direction. You win!

Explanation:

3 0

3 years ago

Which statement best compares momentum and kinetic energy? If you double the velocity of an object, its kinetic energy doubles.

lakkis [162]

If you double the velocity of an object, its momentum doubles. but for the same increase in velocity, the kinetic energy increases 4 times..

7 0

3 years ago

Read 2 more answers

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude

kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

$v_0=115 m/s$ (toward east)

So, decomposing along the x- and y- directions:

$v_{x0} = 115 m/s\\v_{y0} = 0$

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

$a=2.88 m/s^2$ (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

$a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2$

So the two components of the velocity after a time t = 25.0 s will be

$v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s$

So, the magnitude of the velocity of the plane will be

$v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s$

6 0

3 years ago