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4 years ago

Can someone solve this problem and explain to me how you got it

Physics

1 answer:

evablogger [386]4 years ago

4 0

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

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Answer: Option C.

<u>Explanation:</u>

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4 years ago

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform e

nalin [4]

Answer:

$\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

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Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

$r = \frac{0.03}{2}$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^{-4 } \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

$\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

$\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0$

=> $\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }$

=> $\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }$

=> $\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

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3 years ago

Final velocity for 2.6 seconds

enyata [817]

That's going to depend on the initial velocity, and how it changes
or doesn't change during the 2.6 seconds. I've read the question,
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4 years ago

Explain the energy changes involved when a positive charge moves because of a nearby, negatively charged object. use the terms e

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The energy changes involved when a positive charge moves because of a nearby, negatively charged object because that is actually similar to when an object falls in a gravitational field, the potential energy of the object will turn in to a kinetic energy. thank you for this question.

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3 years ago

Read 2 more answers

angela uses a force of 25 newtons to lift her grocery bag while doing 50 joules of work. how far did she lift the grocery bags

Evgen [1.6K]

We know that:
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s=50J/25N
s=2m

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3 years ago

Can someone solve this problem and explain to me how you got it​

Can someone solve this problem and explain to me how you got it