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3 years ago

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FREE BRAINLEIAST FOR FIRST GOGOGOGO!!!!heererer

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Kamila [148]3 years ago

6 0

Answer:

ME PLS

Explanation:

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A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

sladkih [1.3K]

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0

4 years ago

A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha

umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

$R_{1} = 330\ohm$

$R_{2} = 470\ohm$

$R_{3} = 220\ohm$

Since, $R_{2} = 470\ohm$ and $R_{3} = 220\ohm$ are in parallel and this combination is in series with $R_{1} = 330\ohm$ , so,

Equivalent resistance of the circuit is given by:

$R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}$

$R_{eq} = \frac{470\times 220}{470 + 220} + 330$

$R_{eq} = 149.85 + 330 = 479.85 \ohm$

power drain on an ideal battery, P = $\frac{V^{2}}{R_{eq}}$

P = $\frac{2.9^{2}}{479.85}$

P = 0.017 W

4 0

4 years ago

To fight a bacterial infection, a patient was given an antibiotic to take for 10 days. After the patient finished taking the ant

erik [133]

Answer:

ccfvvv

Explanation:vvvvvvvv

3 0

3 years ago

Read 2 more answers

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

LenaWriter [7]

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of <em>free fall objects</em>

Our <u>knowable variables</u> are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the <u><em>rock hits the ground</em></u> making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we <em>evaluate the equation</em>:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of <u><em>Quadratic Formula</em></u>, we can solve it using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the <u><em>time can not be negative</em></u>, the <em>reasonable answer</em> is

$t=6.4534s$

8 0

3 years ago

How is the ideal mechanical advantage of a wheel and axel calculated

kenny6666 [7]

Answer:

by being calculated <em>like duh</em>

Explanation:

5 0

4 years ago

Read 2 more answers