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3 years ago

What is seafloor spreading

Physics

1 answer:

liubo4ka [24]3 years ago

7 0

Answer:

Explanation:

The process by which molten materials add new oceanic crust to the ocean floor.

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Find the work performed when the given force f f is applied to an object, whose resulting motion is represented by the displacem

Ahat [919]

work = force x distance

fd foot pounds

3 0

3 years ago

A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.

kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight= $\sqrt{\frac{2H}{g}}$

substituting value

= $\sqrt{\frac{2*25}{9.8}}$
=2.26seconds

<h3>the ball is in flight before striking the ground for 2.26seconds.</h3>

b. How far from the building does the ball strike the ground?

Horizontal range=?

we have

Horizontal range=u* $\sqrt{\frac{2H}{g}}$

=8.25*2.26
=18.63m

<h3>The ball strikes 18.63m far from building. </h3>

7 0

2 years ago

A:10i - 2j -4k and B: i +7j - k. Determine |A-B|

maksim [4K]

A - B = (10i - 2j - 4k) - (i + 7j - k)

A - B = 9i - 9j - 3k

|A - B| = √(9² + (-9)² + (-3)²) = √189 = 3√19

8 0

2 years ago

A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat

Tanzania [10]

If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 $\text{m/s}^{2}$ .

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is $\omega$ =5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, $a=\omega ^2 r$ .

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

$\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$