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Anvisha [2.4K]
3 years ago
6

Can a particle move in a direction of increasing electric potential, yet have its electric potential energy decrease? Explain

Physics
1 answer:
yan [13]3 years ago
3 0

Answer:

Explanation:

Yes , it is possible .

When a negative charge moves towards a positive charge , it is moving in the direction of increasing electrical potential . In the whole process , its electrical potential energy decreases and its kinetic energy increases .

Actually the potential energy of a negative charge near a positive charge is negative and it is inversely proportional to distance .

V = - Qq / R , When R decreases , the negative value of potential increases . That means potential energy decreases .

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A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th
Igoryamba

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

5 0
3 years ago
The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton
Brrunno [24]

Answer:

A)   F_g = 4.05 10⁻⁴⁷ N, B)   F_e = 9.2 10⁻⁸N, C)    \frac{F_e}{F_g} = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

            F = G \frac{m_1m_2}{r^2}

let's calculate

            F = 6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }

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B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

            F = k \  \frac{q_1q_2}{r^2}

let's calculate

            F = 9 \ 10^9 \  \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}

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C) It is asked to find the relationship between these forces

        \frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }

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therefore the electric force is much greater than the gravitational force

4 0
3 years ago
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expeople1 [14]
Just read the back of the cake box
8 0
4 years ago
Hen a gfci receptacle device is installed on a 20-ampere branch circuit (12 awg copper), what is the minimum volume allowance (i
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Answer:

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How do the positions of the sun and moon affect what people do?
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6 0
3 years ago
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