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Mamont248 [21]
2 years ago
10

Calculate the potential energy of a +1.0μC point charge sitting 0.1m from a -5.0μC point charge.

Physics
1 answer:
AfilCa [17]2 years ago
8 0

Answer:

P.E.      =   -0.449 J

Explanation:

Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.

Now Potential energy is defined by this formula,

P.E. = k q₁ q₂/ r

where P.E. is the potential energy.

k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)

q₁ = charge of one particle = +1.0μC

q₂ = charge of another particle = -5.0μC

r = distance = 0.1 m

Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m

          P.E.      =  -0.449 J

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The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.
Otrada [13]

Answer:

a)   a = 3.06 10¹⁵ m / s , b)    F= 1.43  10⁻¹⁰ N, c)    F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

          I = ∫ F dt = Δp

          F t = m v_{f} - m v₀

          F = m (v_{f} -v₀o) / t

 in the exercise indicates that the speed module is the same, but in the opposite direction

          F = m (-2v) / t

if we use Newton's second law

          F = m a

we substitute

            - 2 mv / t = m a

            a = - 2 v / t

let's calculate

            a = - 2 4.59 10²/3 10⁻¹³

            a = 3.06 10¹⁵ m / s

b)      F= m a

        F= 4.68 10⁻²⁶ 3.06 10¹⁵

        F= 1.43  10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

            F_total = n F

            F_total = n m a

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            F_total = 14.32 10⁻²⁶ N

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3 years ago
Which of the following is not a part of dalton s atomic theory?
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<span>c. atoms are always in motion..............</span>
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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

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or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

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2(V_1+9.0)^2=(2V_1+9.0)^2

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or

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(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

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(2V_1-\sqrt{2}V_1)=(12.72-9.0)

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Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

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