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gizmo_the_mogwai [7]
3 years ago
6

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

Physics
1 answer:
-BARSIC- [3]3 years ago
8 0

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}

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= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

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1181.58+W=1813.05

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3 0
3 years ago
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A penny is dropped off the top of the Stratosphere from rest and falls freely with the
Roman55 [17]

Answer:

S = 122.5m

Explanation:

Given the following data;

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Time, t = 5 seconds

Since it's a free fall, initial velocity, u = 0

To find the displacement, we would use the second equation of motion given by the formula;

S = ut + \frac {1}{2}at^{2}

Where;

  • S represents the displacement or height measured in meters.
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  • t represents the time measured in seconds.
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Substituting into the equation, we have;

S = 0*5 + \frac {1}{2}*(9.8)*5^{2}

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3 0
3 years ago
The force of attraction between a -130.0 C and +180.0 C charge is 8.00 N. What is the separation between these two charges in me
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Answer with Explanation:

The force of attraction between 2 charges of magnitudeq_1,q_2 separated by a distance 'r' is given by

F=\frac{1}{4\pi \epsilon _o}\frac{q_1\times q_2}{r^2}=k\frac{q_1\times q_2}{r^2}

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\epsilon _o is a constant known as permitivity of free space

k=9\times 10^{9}Nm^2/C^2

Applying the given values in the above relation we get

8=9\times 10^{9}\times \frac{130\times 180}{r^{2}}\\\\r^{2}=\frac{9\times 10^{9}\times 130\times 180}{8}\\\\r^{2}=26325\times 10^{9}\\\\\therefore r=\sqrt{26325\times 10^{9}}=5.131\times 10^{6}meters

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C. The bowling ball and the bicycle

p = mv
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