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statuscvo [17]
3 years ago
11

Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f

orm a diamond shape centered on the origin. A charge of +5 μC is placed at the origin. If the force between a +2 μC and a +5 μC charge separated by 10 cm has a magnitude of 9 N, which of the following can we say about the force on the +5 μC charge at the origin in this case?
Physics
1 answer:
mylen [45]3 years ago
7 0

Answer:

The force will be zero

Explanation:

Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.

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3 years ago
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A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are
slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given that,

Mass m = 1.81\times10^{-3}\ kg

Velocity v = (3.00\times10^{4}\ m/s)j

Charge q = 1.22\times10^{-8}\ C

Magnetic field B= (1.63\hat{i}+0.980\hat{j})\ T

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Formula of the acceleration is defined as

F = ma=q(v\times B)

a =\dfrac{q(v\times B)}{m}

We need to calculate the value of v\times B

v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})

v\times B=4.89\times10^{4}

Now, put the all values into the acceleration 's formula

a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}

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7 0
3 years ago
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A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are
zhannawk [14.2K]

Answer:

y = 17.04 ft

Explanation:

Since the cable touches the road at the mid point of two towers

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now we have

y = (4.26 \times 10^{-4})x^2

now we need to find the height at distance of 200 ft from center

so we have

y = (4.26 \times 10^{-4})(200^2)

y = 17.04 ft

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dezoksy [38]

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