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Aleonysh [2.5K]
3 years ago
9

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

rners of a 0.41-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.
Physics
1 answer:
mars1129 [50]3 years ago
3 0

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

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7 0
3 years ago
Determine the resultant moment of the forces acting about the x, y, and z axes. Suppose that f1 = 200 n, f2 = 60 n, f3 = 115 n,
SpyIntel [72]

Based on the forces acting on the axes, the resultant moments will be (345, 400, 600 N·m)

<h3>What would be resultant moment about x-axis?</h3>

= F₃ x 3

= -115 x 3

= -345 N·m

<h3>What would be resultant moment about y-axis?</h3>

= F₁ x 2

= -200 x 2

= -400 N·m

<h3>What would be the resultant moment about z-axis?</h3>

= F₄ x 2

= -300 x 2

= - 600 N·m

In conclusion, the resultant moment about x, y, and z axes is (345, 400, 600 N·m)

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5 0
2 years ago
A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and c
rjkz [21]

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

4 0
3 years ago
A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
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Answer:

628.022466 N

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Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
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