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Natalija [7]
3 years ago
6

Two boxes, P and Q, are at rest on a frictionless horizontal surface. A light, flexible cord connects them. The mass of P is gre

ater than the mass of Q.
(As shown in the diagram,) A horizontal force of magnitude F is applied to box Q, causing both boxes to accelerate to the right.
What best describes the magnitude of the force exerted by the connecting cord on box P?

*equal to F
*greater than F
*zero
*less than F, but greater than zero
Physics
2 answers:
weqwewe [10]3 years ago
8 0

The boxes are tied together, so their motions are equal ... displacement, velocity, and acceleration ... the same for both boxes once they get going.

The force (F) exerted on Q has to be enough to accelerate the mass of both boxes.

But the force of the string pulling box-P is only enough to move the mass of box-P with the same acceleration.

So the magnitude of the force exerted on box-P by the string (the tension in the string) is <em>less than F , (but greater than zero</em>, because it accelerates box-P).

It doesn't matter which box is heavier, lighter, more mass, or less mass.

elixir [45]3 years ago
4 0

the answer is less than F, but greater than zero


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The reason being that only of the x or y component of the field can be cancelled out in either case still the total field can't be reduced to zero.

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-10.8°, or 10.8° below the +x axis

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The y component of the resultant vector is:

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Therefore, the angle between the resultant vector and the +x axis is:

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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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Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

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

\displaystyle F=\frac{k\ q_1\ q_2}{r^2}

Where k is the proportional constant of value

k=9*10^9\ N.m^2/c^2

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

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

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

The force between charges separated 2d is

\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}

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

And the force between the charges A and D is

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

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

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

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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bearhunter [10]

Answer: 0 NEWTONS

Explanation:

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