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marissa [1.9K]
3 years ago
12

A charge moves from pount A to point B in an electric field. what is true about the potential energy of the charge?

Physics
2 answers:
devlian [24]3 years ago
6 0

Answer:

B) it is dependent on the value of the charge and its position

Explanation:

As we know that electrostatic potential energy is given by the formula

U = \frac{kq_1q_2}{r}

here we know that electric potential due to one charge on the position of other charge is given by the equation

V = \frac{kq_1}{r}

so potential energy can be given by the formula

U = q_2 V

so here potential of a given position will be different so it is dependent on the position of charge

also potential energy depends on the magnitude of charge

so correct answer will be

B) it is dependent on the value of the charge and its position

iren [92.7K]3 years ago
3 0
I'm going to have to say its B
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You are the juror of a case involving a drunken driver whose 1041 kg sports car ran into a stationary 1928 kg station wagon stop
mina [271]
1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g 
2. Find the acceleration due to friction: 
F = m*a =  μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t² 
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:

m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash

m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁ 
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s


7 0
3 years ago
Particles q1, 92, and q3 are in a straight line.
NNADVOKAT [17]

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}}  = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N

The force,by the charge q₂ on the q₃;

\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}}  = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09  \ N

The net force is the sum of the two forces;

\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

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goldenfox [79]
Increases is the correct answer
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xz_007 [3.2K]

Answer:

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Explanation:

I just took the test, Hope it helps!

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1.) dependent variable A.)the variable representing the
My name is Ann [436]

Answer:

Explanation:

the answer is b

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