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Andre45 [30]
4 years ago
11

At what position or positions on the x-axis is the electric field zero?

Physics
1 answer:
ElenaW [278]4 years ago
4 0

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

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A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of
lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight  (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0  

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

 m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W=  ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7  N

8 0
3 years ago
How is Newtons second law apply to the egg drop ?
kotykmax [81]

Answer:

Newton's Second Law is applied because of the acceleration caused by natural forces as the egg is plummeting to the earth. And the amount of acceleration the egg has will be largely affected by the amount of force Mr. Baker uses to hurl the egg to the ground.

Explanation:

hope that helps

3 0
3 years ago
Two blocks A and B have a weight of 11 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of
Andreas93 [3]

Answer:

the block that starts moving first is block A ,    fr = 1.625 N ,  fr = 1.5 N

Explanation:

For this exercise we use Newton's second law, for which we take a reference system with the x axis parallel to the plane and the y axis perpendicular to the plane

X axis

       fr- Wₓ = 0

       fr = Wₓ

Axis y

      N- W_{y} = 0

      N = W_{y}

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     Cos θ = W_{y} / W

     Wₓ = W sin θ

     W_{y} = W cos θ

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     W_{y} = 11 cos θ

The equation for friction force is

      fr = μ N

   

We substitute

      μ (W cos θ) = W sin θ

      μ = tan θ

We can see that the system began to move the angle.

         θ = tan⁻¹ μ

So the angles are

Block A      θ = tan⁻¹ 0.15

           θ = 8.5º

Block B      θ = tan⁻¹ 0.26

             θ = 14.6º

So the block that starts moving first is block A

The friction force is

         

Block A

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         fr = 1.625 N

Block B

         fr = 6 sin 14.6

         fr = 1.5 N

5 0
3 years ago
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Elena-2011 [213]

Answer:

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f=uv/(u+v)

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f=10.5cm

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Gold has 118 neutrons. If a gold atom were to lose a neutron, what would happen?
nexus9112 [7]

Answer:

answer is B! if it adds one it becomes platinum and if it loses one it becomes mercury.

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