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bulgar [2K]
3 years ago
6

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

er opposite corners. Determine the magnitude and direction of the force on right lower corner. Please explain.

Physics
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

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olga_2 [115]
<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force F_{r} is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma    -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F -  F_{r}

m = 50kg

a = 0   [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F -  F_{r} = m x a

F -  F_{r} = 50 x 0

F -  F_{r} = 0

F =  F_{r}            -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force F_{r} is opposing the movement of the box.

∑F = 1.5F -  F_{r}

m = 50kg

a =  0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F -  F_{r} = m x a

1.5F -  F_{r} = 50 x 0.1

1.5F -  F_{r} = 5            ---------------------(iii)

<em>Substitute </em>F_{r}<em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5            

0.5F = 5            

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

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