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Vikki [24]
4 years ago
14

An airplane flies eastward and always accelerates at a constant rate. at one position along its path it has a velocity of 26.3 m

/s, it then flies a further distance of 48100 m and afterwards its velocity is 41.9 m/s. find the airplane\'s acceleration and calculate how much time elapses while the airplane covers those 48100 m.
Physics
1 answer:
galina1969 [7]4 years ago
6 0
We will use the formula / equation to determined the time.

Distance = ½ * (vi + vf) * t 
48100 = ½ * (26.3 + 41.9) * t 
t = 48100 ÷ 34.1 = 1410.557185 seconds 

We will use the formula / equation to determined the acceleration. 

vf = vi + a * t 
41.9 = 26.3 + a * 1410.557185 
a = (41.9 – 26.3) ÷ 1410.557185 = 0.011059459 m/s^2 

We will use the formula / equation to determined the acceleration.

vf^2 = vi^2 + 2 * a * d 
41.9^1 = 26.3^2 + 2 * a * 48100 
a = (41.9^2 – 26.3^2) ÷ 96200 = 0. 011059459 m/s^2 
Since both answers are the same, I believe the acceleration is correct.
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Answer:

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Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

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