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Ilia_Sergeevich [38]
3 years ago
14

Two positive point charges with charges q1 and q2 are separate by a distance of 30cm. A third positive charge qo is placed betwe

en the two charges at 8cm on the right of the charge q1. See picture. The charge qo is equal to 3 µC (micro-coulombs = 10-6 C). The Coulomb force exerted by q1 on qo is 25 N and the Coulomb force exerted by q2 on qo is 10 N. Determine the net electric field in magnitude and direction on the charge qo due to the other two charges.
Physics
1 answer:
vivado [14]3 years ago
5 0

Answer:

The magnitude of net electric is 5\times 10^6\ \rm N/C from charge q_1 to q_2 along the line joining them.

Explanation:

Given:

Distance between the charges =30 cm

Magnitude of charge q_0 =3\ \rm \mu C

Force exerted by q_1 on q_0=25 N

Force exerted by q_2 on q_0=10 N

Now according to coulombs Law we have

\dfrac{kq_1q_0}{0.08^2}=25\\\\q_1=59.25\times10^{-7}\ \rm C

similarly

\dfrac{kq_2q_0}{0.22^2}=10\\\\q_2=179.25\times10^{-7}\ \rm C

Noe the electric field at eh position of charge q_0 is given by

Let E_1  be the electric field due to charge  q_1` and  E_1 bet he electric field due to charge  q_2`

then The net electric Field at the point is given by

E_{net}=E_1-E_2\\=\dfrac{kq_1}{0.08^2}-\dfrac{kq_2}{0.22^2}\\\\=\dfrac{9\times10^9\times59.25\times10^{-7}}{0.08^2}-\dfrac{9\times 10^9\times 179.25\times 10^{-7}}{0.22^2}\\\\=5\times 10^6\rm N/C

The direction of electric Field is from charge q_1 to q_2 along the line joining them.

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A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is t
GuDViN [60]

Answer:

The magnification is  m =  12

Explanation:

From the question  we are told that

   The object distance is u  = 36.2 \ cm

     The focal length is  v  =  39.5 \ cm

From the lens equation we have that

         \frac{1}{f}  =  \frac{1}{u} +  \frac{1}{v}

=>     \frac{1}{v}  =  \frac{1}{f}  - \frac{1}{u}

substituting values

       \frac{1}{v}  =  \frac{1}{39.5}  - \frac{1}{36.2}

       \frac{1}{v}  =  -0.0023

=>   v =  \frac{1}{0.0023}

=>   v =-433.3 \ cm

The magnification is mathematically represented as

         m =-   \frac{v}{u}

substituting values

        m =-   \frac{-433.3}{36.2}

         m =  12

         

6 0
2 years ago
List and define three types of intermolecular forces and identify which types of molecules each forces affects.
Zigmanuir [339]

Van der waals forces

Hydrogen bonding

Crystal lattice forces

Explanation:

Intermolecular forces exists between molecules.

  • Van der waals forces are weak attractions that joins non-polar and polar molecules together. London dispersion forces occurs between non-polar molecules(polar) and noble gases. Dipole -dipole attraction occurs between polar molecules. Van der waal forces occurs in graphite layers, HCl e.t.c
  • Hydrogen bonding is force of attraction between polar molecules in which a hydrogen atom is directly joined to a highly electronegative atom. Examples occur in water.
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Learn more:

Intermolecular forces brainly.com/question/3622116

#learnwithBrainly

3 0
3 years ago
Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinal and transverse modul
gregori [183]

Answer:

It is not possible to produce continued and oriented fiber.

Explanation:

To solve the problem it is necessary to take into account the concepts related to Fiber volume ratio. The amount of fiber in a fiber reinforced compound corresponds directly to the mechanical properties of the compound. Given the fiber volume fraction, the theoretical elastic properties of a compound can be determined. The elastic modulus of a compound in the fiber direction of a unidirectional compound can be calculated using the following equation:

E = (1-V_f)E_m+V_fE_f

Where,

E is the longitudinal modulus of Elasticity

V_f is the fiber volume ratio

E_m is the elastic modulus of the matrix

E_f is the elastic modulus of the fibers

We need to consult the table of characteristics of Fibers and Reinforcements of Materials, in which they specify that the modulus of elasticity of the aramid fiber-epoxy is

E_f = 131Gpa

Moreover from the statement,

E = 35Gpa

E_m = 3.4Gpa

Replacing in the previous equation,

35 = 3.4 (1-V_f)+131V_f

V_f = 0.25 \rightarrow longitudinal

To make the comparison we now calculate the Fiber volume ratio through the transverse elastic modulus,

E = \frac{E_mE_f}{(1-V_f)E_f+V_fE_f}

Our values are given in this case as:

E = 5.17Gpa\\E_m = 3.4Gpa \\E_f = 131Gpa

Replacing,

5.17 = \frac{3.4*131}{(1-V_f)(131)+V_f*3.4}

V_f = 0.351 \rightarrow transversal

From both cases it is possible to conclude that it is not possible to produce a fiber of the specified material in a continuous and oriented manner, as long as the volume fraction is different in the different cases.

5 0
3 years ago
In which task is a camera the most useful?
kupik [55]

Answer:

B, to predict star count

Explanation:

the rest can be done with rulers, Einstein cans and public surveys, we can't predict how many stars there will be, but we can take pictures of them instead

8 0
2 years ago
Mike Powell holds the record for the long jump of 8.95 m, established in 1991. If he left the ground at an angle of 18.8°, what
maksim [4K]

Answer:

<em>12 m/s</em>

Explanation:

<u>Projectile Motion</u>

It's also known as 2D motion because the movement takes place in both axis x and y. The x-axis motion is at a constant speed since in absence of friction, no external force stops or accelerates the object. The y-axis motion is at variable speed, which is changed by the acceleration of gravity that makes the object to reach a maximum height and then go back to ground level.

The maximum horizontal distance reached (also called Range) is given by

\displaystyle X_m=\frac{V_{o}^2sin2\theta}{g}

Knowing that \theta=18.8^o, X_m=8.95\ m, we solve for Vo

\displaystyle V_o=\sqrt{\frac{gX_m}{sin2\theta}}

\displaystyle V_o=\sqrt{\frac{9.8\cdot 8.95}{sin(2\cdot 18.8^o)}}=12\ m/s

Thus, the initial speed of Mike Powell was 12 m/s

7 0
3 years ago
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