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MAVERICK [17]
2 years ago
9

A charge of 6. 4 × 10–7 C experiences an electric force of 1. 8 × 10–1 N. What is the electric field strength? 1. 1 × 10–7 N/C 3

. 6 × 10–6 N/C 8. 2 × 100 N/C 2. 8 × 105 N/C.
Physics
1 answer:
shutvik [7]2 years ago
3 0

The electric field strength is defined as the ratio of electric force and charge. The electric field strength will be 2.8 ×10⁵ N/C.

<h3>What is electric field strength?</h3>

The electric field strength is defined as the ratio of electric force and charge. The field's direction is determined by the direction of the force acting on the positive charge.

A positive charge produces a radially outward electric field, whereas a negative charge produces a radially inward electric field.

The given data in the problem is;

Q is the charge= 6. 4 × 10–7 C

E is the electric force=1. 8 × 10–1 N

The relation between the electrostatic force and the electric field due to a charge is given as:

\rm F_E=Q E \\\\ \rm E=\frac{F_E}{Q}  \\\\  \rm E=\frac{1.8\times 10^{-1}}{6.4\times 10^{-7}}  \\\\ \rm E=2. 8 \times 10^5\ N/C.

Hence the electric field strength will be 2.8 ×10⁵ N/C.

To learn more about the electric field strength refer to the link;

brainly.com/question/4264413

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Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

      $=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

    $= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

    = 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

                                           $ =\frac{1}{2} \times 1.225  \times 200^2$

                                          $=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

     $=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

     = 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

                  $=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

                  = 270 N

Therefore the net drag = 270 x 2

                                      = 540 N

7 0
2 years ago
An object is moving at a constant velocity of 10 m/s for 5 seconds. What is
german

initial velocity (u)=0m/s

final velocity (v)=10m/s

time( t)=5s

acceleration (a)=v-u÷t

acceleration (a)=10-0÷5

acceleration (a)=10÷5

acceleration (a)=2

therefore acceleration (a)=2m/s

3 0
3 years ago
Carla draws two circuit diagrams that connect the same components in different ways, as shown. which statement about the circuit
vagabundo [1.1K]

Carla draws two circuit diagrams that connect the same components in different ways, as shown. The correct statement is "the total resistance in circuit a is greater than that in-circuit b.". Option A. This is further explained below.

<h3>What is a circuit?</h3>

Generally, a circuit is simply defined as a fixed schedule of activities or places for a certain activity, usually including athletics or public performance.

In conclusion, Carla creates two circuit schematics, as illustrated, that link the same components in various ways. "The overall resistance in circuit an is larger than that in b," is the correct statement.

Read more about  resistance

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5 0
2 years ago
What is the charge on a
Marrrta [24]

Answer:

positive

Explanation:

The electrons revolves around the nucleus and they are negatively charged. The nucleus consists of protons and neutrons. The neurons have no charge but the protons are positively charged.

7 0
3 years ago
Consult Multiple-Concept Example 5 to review the concepts on which this problem depends. A light bulb emits light uniformly in a
den301095 [7]

Answer:

a. 0.332 W/m² b. 11.2 V/m c. 15.8 V/m

Explanation:

(a) the average intensity of the light,

Intensity, I = P/A where P = average power = 150.0 W and A = area through which the power emits = 4πr² where r = distance from bulb = 6 m.

So, I = P/A = P/4πr²

Substituting the values of the variables into the equation, we have

I =  P/4πr²

I =  150.0 W/4π(6 m)²

I =  150.0 W/4π(36 m²)

I =  150.0 W/452.39 m²

I =  0.332 W/m²

(b) the rms value of the electric field,

Since Intensity, I = E²/cμ₀ where E = rms value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m.

Making E subject of the formula, we have

E² = Icμ₀

E = √(Icμ₀)

Since I = 0.332 W/m², substituting the other terms into the equation, we have

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(12.5 × 10)

E = √125 V/m

E = 11.18 V/m

E ≅ 11.2 V/m

(c) the peak value of the electric field.

The peak value of electric field, E' is gotten from E = E'/√2 where E = rms value of electric field.

So, E' = √2E

= √2 × 11.2 V/m

= 15.81 V/m

≅ 15.8 V/m

6 0
3 years ago
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