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inna [77]
4 years ago
10

Consider a pair of charged particles, particles 1 and 2, separated by a distance r. 6) Which options below are true? [Enter your

answer as a string of the letter options you believe are correct. For instance, if you think options A, D, and E are correct, then enter "ADE"] A) Particle 2's electric field contributes to the electric force on particle 1. B) Particle 1's electric field does not contribute to the electric force on particle 1. C) If particle 2 were removed, there would still be a non-zero electric field at its location. D) If particle 2's distance from particle 1 is tripled, particle 1's electric field at 2's new location is no different than before since the field does not depend on the the properties of particle 2. E) None of the above
Physics
1 answer:
polet [3.4K]4 years ago
6 0

Answer:

A, B and C

Explanation:

Option A is correct because, the force on particle 1 depends on the electric field due to particle 2. Since the electric field due to particle 2 is the force per unit charge acting at the point of particle 1, the force at the point of particle 1 is the product of that electric field due to particle 2 and the charge on particle 1. So, Particle 2's electric field contributes to the electric force on particle 1.

Option B is correct because the force on a charge can only be exerted due to an external electric field. So, Particle 1's electric field does not contribute to the electric force on particle 1

Option C is correct since an electric field is always present in the region surrounding a charge irrespective of whether there is a charge in the region of the electric field or not. So, If particle 2 were removed, there would still be a non-zero electric field at its location.

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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm . The explorer finds that
iren2701 [21]

Answer: g = 10.0 m/s/s

Explanation:

For a simple pendulum, provided that the angle between the lowest and highest point of  his trajectory be small, the oscillation period is given by the following expression:

T = 2π √L/g , where L = pendulum length, g= accelleration of gravity.

We can also define the period, as the time needed to complete a full swing, so from the measured  values, we can conclude the following :

T = 140 sec/ 101 cycles = 1.39 sec

Equating both definitions for T, we can solve for g, as follows:

g = 4 π² L / T² = 4π². 0.49 m / (1.39)² = 10.0 m/s/s

7 0
3 years ago
You've been tired and lethargy what are two possible reasons?
dybincka [34]

Explanation:

physical exertion.

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lack of sleep.

being overweight or obese.

periods of emotional stress.

boredom.

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3 years ago
Tragically, the united states lost astronauts in accidents involvin
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The quest to put Americans on the moon before the Soviets do. During the Space Race, a couple of astronauts went up into space only to be tragically killed in an explosion. 
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According to the kinetic molecular therory, the pressure of a gas in a container will increase if the
Masja [62]

According to the kinetic molecular theory, the pressure of a gas in a container will increase if the number of collisions with the container wall increases.

<u>Explanation:</u>

Keeping the volume of the vessel constant, if we increase the amount of gas in it; the pressure will increase. This is because when the number of gas particles increases in that limited volume, they hit the walls of the container with more energy and hence, the overall pressure of the gas increases.

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5 0
4 years ago
Can someone solve this problem and explain to me how you got it​
evablogger [386]

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

4 0
4 years ago
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