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Alexxandr [17]
3 years ago
13

URGENT!!!!!!! PLEASE HELP WITH THIS PHYSICS PROBLEM

Physics
1 answer:
Levart [38]3 years ago
3 0

Explanation:

Let

x_1 = distance traveled while accelerating

x_2 = distance traveled while decelerating

The distance traveled while accelerating is given by

x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2

\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2

\:\:\:\:\:= 1125\:\text{m}

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

v^2 = v_0^2 + 2ax_2

0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2

Solving for x_2, we get

x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}

\:\:\:\:\:= 4327\:\text{m}

Therefore, the total distance x is

x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}

\:\:\:\:= 5452\:\text{m}

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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o
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Two point-charges q_1 and q_2 separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

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\displaystyle F_1=\frac{k\ Q\ Q}{d^2}

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\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}

\displaystyle F_3=\frac{k\ Q^2}{9d^2}

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Charge A. It receives force to the left from B and C and to the right from D

\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_A=-\frac{41}{36}F_1

Charge B. It receives force to the right from A and D and to the left from C

\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}

\displaystyle F_B=\frac{1}{4}F_1

Charge C. It receives forces to the right from all charges.

\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}

\displaystyle F_C=\frac{9}{4}F_1

Charge D. It receives forces to the left from all charges

\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}

\displaystyle F_D=-\frac{49}{36}F_1

Comparing the magnitudes of each force is just a matter of computing the fractions

\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9

The greatest force is 9 times the smallest net force

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