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3 years ago

100 POINTS PLEASE HELP I NEED A GOOD GRADE DESPERATELY TO PASS!!!

Physics

2 answers:

xxMikexx [17]3 years ago

8 0

Answer:

set A is showing the same position in both graphs

anzhelika [568]3 years ago

5 0

Answer:

Explanation:

The answer is B.

I'm positive it's B

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Cosmologists are scientists who study stars, galaxies, black holes, and other objects in the universe. In order to study the bil

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3 years ago

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Which of the following is not true about the Law of Conservation of Energy?

SVEN [57.7K]

By definition, the law of conservation of energy states that:

Ei = Ef

Where,

Ei: initial energy

Ef: final energy

Therefore, no matter the type of energy, always the final energy is equal to the final energy.

Energy can be transformed into another type of energy. For example, the potential energy can be transformed into kinetic energy.

Also, energy is not created, nor destroyed.

Answer:

The following is not true about the Law of Conservation of Energy:

A. It states that the total energy in the universe keeps increasing.

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3 years ago

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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 t :

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

• position at time t :

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