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Bumek [7]
3 years ago
6

A negatively charged rubber rod can pick up small, neutrally charged pieces of paper by a process called

Physics
1 answer:
Ronch [10]3 years ago
4 0
The answer would be intoduction hope this really helped
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Yes they are necessary 
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B. number of oscillations in a given period of time.
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How Are Babies In The Womb For 9Months But Aren't 9Months When They Are Born
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<span>You could easily define it this way. This would be valid. But for most of history it was unclear when a baby was conceived, so it would be silly for someone to say their baby was 9 months old when it was actually born at only 8.5 months. Days of conception similarly would not be a good substitute for birthdays because no one would know them. Then in general it was much easier to mark someone's age as when they came into the world as an independent being, and this tradition is far too entrenched (and simple and easy to handle and universal for people in poor countries etc) to think we would at all gain from now finding exactly the date of conception and counting age from that point.</span>
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At t= 0, a particle moving in the xy plane with constant acceleration has a velocity of Vi= (3.00i -2.00j) m/s and is at the ori
ivanzaharov [21]

Answer:

(a) a = (2i + 4.5j) m/s^2

(b) r = ro + vot + (1/2)at^2

Explanation:

(a) The acceleration of the particle is given by:

\vec{a}=\frac{\vec{v}-\vec{v_o}}{t}\\\\

vo: initial velocity = (3.00i -2.00j) m/s

v: final velocity = (9.00i + 7.00j) m/s

t = 3s

by replacing the values of the vectors and time you obtain:

\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2

(b) The position vector is given by:

\vec{r}=\vec{r_o}+\vec{v_o}t+\frac{1}{2}\vec{a}t^2

where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2

4 0
3 years ago
The oscillating current in an electrical circuit is as follows, where I is measured in amperes and t is measured in seconds. I =
Vesna [10]

Answer:

Explanation:

Given

I=7\sin (60\pi t)+\cos (120\pi t)

=\frac{1}{\frac{1}{60}}\times \int_{0}^{\frac{1}{60}}7\sin (60\pi t)dt+\int_{0}^{\frac{1}{60}}\cos (120\pi t)dt

=60\left [ \frac{-7}{60\pi }\cos (60\pi t)+\frac{1}{120\pi }\sin (120\pi t)\right ]_{0}^{\frac{1}{60}}

=60\left [ \frac{-7}{60\pi }\left ( \cos \pi-\cos 0\right )+\frac{1}{120\pi }\left ( \sin (2\pi )-\sin (0)\right )\right ]

=\frac{-7}{\pi }\left ( -1-1\right )+\frac{1}{2\pi }\left ( 0-0\right )

=\frac{14}{\pi }

3 0
3 years ago
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