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3 years ago

13

How much force is needed to accelerate a 9760 kg airplane at a rate of 4.6 m/s^2

1 answer:

seropon [69]3 years ago

8 0

Assuming air resistance / friction is negligable:

force = mass × acceleration

f = m × a
f = 9760 × 4.6
f = 44896N

f= 45kN (2 sig fig)

I hope this helps. If you have any questions, feel free to ask

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Explanation:

We are given that

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Distance between two charges, r=2cm= $2\times 10^{-2}$ m

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Sum of two charges= $10\mu C$

Let one charge= $q_1=q\mu C=q\times 10^{-6}C$

$q_2=(10-q)\times 10^{-6} C$

We know that

Electric force between two charges

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

Where $k=\frac{1}{4\pi \epsilon_0}=9\times 10^{9}$

Using the formula

$36\times 10^{-3}=9\times 10^{9}\times \frac{q\times 10^{-6}\times(10-q)\times 10^{-6}}{(2\times 10^{-2})^2}$

$\frac{144\times 10^{-7}}{9\times 10^{9}\times 10^{-12}}=q(10-q)$

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$q^2-10q+0.0016=0$

$10000q^2-100000q+16=0$

$q=\frac{100000\pm\sqrt{(100000)^2-4\times 10000\times 16}}{2\times 10000}$

Using the formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$q=9.999$ and $q=0.00016$

$q_2=10-9.9998=0.0002$

$q_2=10-0.00016=9.99984$

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