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Serhud [2]
3 years ago
5

Help Pleases

Physics
1 answer:
Licemer1 [7]3 years ago
6 0
The answer will be B. INSULATORS

With their strong dielectric properties, silicones are used with cables and cable terminations, corona-resistant insulation tubing, keyboards, and contact mats. Particle-filled silicones that provide EMI shielding are used in conductive seals because metal or metal-coated particles also provide electrical conductivity.

Eat your beans.
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A person jumps from the roof of a house 3.1-m high. When he strikes the ground below, he bends his knees so that his torso decel
Sveta_85 [38]
Part A

Free fall motion
h = 3.1 m

Equation: Vf = √(2gh) = √(2*9.8 m/s^2 * 3.1 m) = 7.8 m/s

That is the only part in the question.
3 0
4 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
Help please! This question is driving me crazy
Varvara68 [4.7K]

Answer:

-10.8°, or 10.8° below the +x axis

Explanation:

The x component of the resultant vector is:

x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)

x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

θ = atan(y / x)

θ = atan(-0.777 / 4.07)

θ = -10.8°

The angle is -10.8°, or 10.8° below the +x axis.

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The amount of time required for 2 successive wave crests to pass a fixed point is called wave ________.
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I believe this is known as wave period.
hope this helps!
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