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kompoz
10 days ago
13

Two charges q1 = 5 µC, q2 = -26 µC, are L = 19 cm apart. A third charge is to be placed on the line between the two charges. How

far from q1 should the third charge be placed so that the net electric force on the third charge is minimized?
Physics
1 answer:
Keith_Richards [3.1K]10 days ago
6 0
The electric force between two objects is expressed as being proportional to the product of their charges and inversely proportional to the square of the distance separating them. In this instance, the distance between the first two charges is 19 cm. We formulate the equation k q1 q3/ (x)^2 = k q2 q3/ (19-x)^2, where x denotes the separation between q1 and q3. The charge q3 cancels out, and q2 is used in absolute terms. The resulting value of x is 5.79 cm.
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The rotational speeds of four generators are listed in RPM (revolutions per minute). Arrange the generators in order based on th
Keith_Richards [3153]

Utilizing the same generator, the only variable affecting the electric field is the speed. The higher the generator's rotation speed, the more extensive the electric field it generates. Hence, the order is 3000 rpm<3200 rpm<3400 rpm<3600 rpm

4 0
17 days ago
Read 2 more answers
Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the
Yuliya22 [3234]

Response:

The acceleration of car 2 is four times that of car 1.

Rationale:

Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:

a=\dfrac{v^2}{r}

In this scenario, two race cars are moving at consistent speeds around a circular course. Both automobiles are located at an equal distance from the center, but car 2 is operating at twice the speed of car 1.

Thus,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}

1 and 2 represent the first and second cars, respectively.

v_2=2v_1

Consequently,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1

Therefore, car 2's acceleration is four times that of car 1.

4 0
1 month ago
A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its no
Ostrovityanka [3092]

Response:

v = 4.08 m/s²

Clarification:

4 0
1 month ago
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Sav [3056]
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = &pi; r^2 = 100 &pi; =314 mm^2
(i) Stress,
&sigma;
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
&epsilon;
= ratio of extension / original length
= &sigma; / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= &epsilon; * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
4 days ago
A wire has an electric field of 6.2 V/m and carries a current density of 2.4 x 108 A/m2. What is its resistivity
Softa [2965]

Response:

The resistivity can be expressed as \rho = 2.5 *10^{-8} \ \Omega \cdot m

Clarification:

According to the information provided,

    The value of the electric field measures  E = 6.2 V/m

     The density of current is given as  J = 2.4 *10^{8} \ A/m^2

Typically, resistivity is represented in mathematical terms as

         \rho = \frac{E}{J}

by inserting values

        \rho = \frac{6.2}{2.4 *10^{8}}

        \rho = 2.5 *10^{-8} \ \Omega \cdot m

5 0
1 day ago
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