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AlekseyPX
2 years ago
13

Four charges are placed on the corners of a rectangle. What is the resultant force on the positive charge (a = 1.1 m, b = 0.9 m,

q = 2.3 × 10-9C)?
Physics
1 answer:
lord [1]2 years ago
4 0

Answer:

F = 93.49 × 10^(-9) N

Explanation:

We are given;

Length of rectangle; a = 1.1 m

Width of rectangle; b = 0.9 m

Charge on rectangle; q = 2.3 × 10^(-9) C

Formula for force is;

F = kq²/r²

Where;

k is a constant = 8.99 x 10^(9) N.m²/C²

a) on the width side, we have;

F = [8.99 x 10^(9) × (2.3 × 10^(-9))²]/0.9²

F = 58.79 × 10^(-9) N

This is on the y-axis

b) on the length side, we have;

F = [8.99 x 10^(9) × (2.3 × 10^(-9))²]/1.1²

F = 39.3 × 10^(-9) N

This is on the x-axis

C) using pythagoras theorem, for the diagonal side, we have c² = a² + b².

Thus;

F = [8.99 x 10^(9) × (2.3 × 10^(-9))²]/(1.1² + 0.9²)

F = 23.54 × 10^(-9) N

Using trigonometric ratios, we can find the angle θ.

tan θ = b/a

tan θ = 0.9/1.1

tan θ = 0.8182

θ = tan^(-1) 0.8182

θ = 39.29°

Resolving along x and y axis, we have;

F_x = 23.54 × 10^(-9) × cos 39.29°

F_x = 18.22 × 10^(-9) N

F_y = 23.54 × 10^(-9) × sin 39.29°

F_y = 14.91 × 10^(-9) N

Resultant force will be;

F = √((Σf_x)² + (ΣF_y)²)

F = √[(18.22 × 10^(-9)) + (39.3 × 10^(-9))]² + [(14.91 × 10^(-9)) + (58.79 × 10^(-9))²]

F = √[(33.0855 × 10^(-16)) + (54.317 × 10^(-16))]

F = 93.49 × 10^(-9) N

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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th
-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = -13.125 m/s^2

Explanation:

By momentum conservation along x direction we will have

mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2

so we have

v_1 + v_2 = 2v

v_1 + v_2 = 4.68

also by energy conservation

\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J

\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17

(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)

(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83

21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83

2v_2^2 - 9.36v_2 + 2.12 = 0

by solving above equation we will have

v_1 = 4.44 m/s

v_2 = 0.24 m/s

Average acceleration on first part of the chunk is given as

a_1 = \frac{4.44 - 2.34}{0.16}

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = \frac{0.24 - 2.34}{0.16}

a_2 = -13.125 m/s^2

5 0
4 years ago
n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
What is it called when the right side of a design is reflected across a central axis and mirrored on the left side of the design
guapka [62]

Answer:

What is it called when the right side of a design is reflected across a central axis and mirrored on the left side of the design?

5 0
3 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

3 0
3 years ago
A solid cube of side 5cm has a mass of 250g. What is its density in kgm ??​
Mariulka [41]

Answer:

5kgm

Explanation:

convert cm to m and g to kg

250/1000=0.25kg

5/1000=0.05m

then find the density

density=mass/volume

=0.25kg/0.05m

=5kgm

7 0
3 years ago
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