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Vinil7 [7]
2 years ago
7

Three point charges are on the x axis: −1 µC

Physics
1 answer:
Nina [5.8K]2 years ago
5 0

Answer:

0.0078 N

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

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

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

The force is attractive if the two  charges have opposite sign, and repulsive if the two charges have same sign.

In this problem, we have:

q_1 = -1 \mu C = -1 \cdot 10^{-6}C located at x_1=-3 m

q_2=+9 \mu C = +9\cdot 10^{-6}C located at x_2=0 m

q_3 = -5 \mu C = -5\cdot 10^{-6} C located at x_3=+3 m

The force between charge 1 and charge 2 is:

F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N

And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.

The force between charge 1 and charge 3 is:

F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N

And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.

Therefore, the net  force on charge 1 is:

F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N

towards the right.

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Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulatin
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Explanation:

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         Charge on block A = \frac{(8.7 + 0 nC}{2}

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Therefore, final charge on the block A is 4.35 nC.

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Thus, we can conclude that while the blocks were in contact with each other then electrons will flow from A to B.

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2 years ago
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A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri
san4es73 [151]

Answer:

T = 8.55 N

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When string makes an angle 40 degree with the vertical then it will have two forces on it

1) gravitational force (mg)

2) Tension force in string (T)

now we know that net force towards the center of the path is known as centripetal force and it is given as

T - mg cos40 = F_c

T - (0.40\times 9.8)cos40 = \frac{mv^2}{L}

T = 3 + \frac{0.40\times 5^2}{1.8}

T = 3 + 5.55

T = 8.55 N

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2 years ago
A 3.5 kg object moving at 8.1 m/s in the positive direction of an x axis has a one-dimensional elastic collision with an object
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Answer:

So the mass of the second object M will be 1.951 kg    

Explanation:

We have given mass of the first object m_1=3.5kg and its velocity v_1=8.1m/sec

Mass of the second object m_2=M  it is at rest so its velocity v_2=0

From conservation of momentum we know that

Initial momentum = final momentum

So m_1v_1+m_2v_2=(m_1+m_2)v

3.5\times 8.1+M\times 0=(3.5+M)5.2

28.35=18.2+5.2M

M = 1.951 kg

8 0
2 years ago
A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
kherson [118]

Answer:

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

5 0
3 years ago
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