What kind of system does not allow matter or energy to enter or exit?

aliaa is skating down a ramp. the wheels on her skateboard have diameters of 60 \text{ mm}60 mm60, start text, space, m, m, end

GarryVolchara [31]

The total distance traveled by Aliaa using her skateboard for 20 revolution of the wheels is equal to 3770 millimeters.

<u>Given the following data:</u>

Diameter of skateboard = 60 mm.

Number of revolution = 20 revolutions.

Radius = diameter/2 = 60/2 = 30 mm.

<h3>What is distance?</h3>

Distance can be defined as the amount of ground covered (traveled) by a physical object over a specific period of time and speed, regardless of its direction, starting point or ending point.

For one revolution of the wheels, the distance traveled by Aliaa using her skateboard is given by:

Distance = 2πr

Distance = 2 × 3.142 × 30

Distance = 188.5 mm.

Therefore, the total distance traveled by Aliaa using her skateboard for 20 revolution of the wheels is given by:

Distance = 188.5 × 20

Distance = 3770 millimeters.

Read more on distance per revolutions here: brainly.com/question/10989073

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2 years ago

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

-BARSIC- [3]

Given the particle's acceleration is

$\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath$

with initial velocity

$\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath$

and starting at the origin, so that

$\vec r(0) = \vec 0$

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

$\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du$

$\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du$

We have

• velocity at time <em>t</em> :

$\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}$

• position at time <em>t</em> :

$\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}$