Suppose you increase your walking speed from 4 m/s to 13 m/s in a period of 3 s. What is your acceleration?

Why did the potassium permanganate crystals start to dissolve in water without being stirred or shaken

KonstantinChe [14]

<span>because the substance in the potassium permanganate crystals are permeable to water, so that means it will dissolve instantly while poured into water </span>

7 0

3 years ago

Read 2 more answers

Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangmen

murzikaleks [220]

Answer:

a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d) W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

W = ΔU = U_∞ -U

the potential energy for point charges is

U =k $\sum \frac{q_i q_j}{r_{ij} }$

in this case we only have two particles

U = k $\frac{q_1q_2}{r_{12} }$

the distance is

r₁₂ = $\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }$

r₁₂ = $\sqrt{ 0 + ( 2-0)^2}$ Ra 0 + (2-0)

r₁₂ = √2= 1.4142 m

we substitute

W = k \sum \frac{q_i q_j}{r_{ij} }

let's calculate

W = $\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}$ 9 109 1.6 10-19 1.6 10-19 / 1.4142

W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

in this case we have two fixed electrons

U = k $( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )$

in this case all charges are electrons

q₁ = q₂ = q₃ = q

W = U = k q² $( \frac{1}{r_{13} } + \frac{1}{r_{23} } )$

the distances are

r₁₃ = $\sqrt{(3-0)^2 + 0}$ RA (3.00 -0) 2 + 0

r₁₃ = 3

r₂₃ = $\sqrt{ 3^2 + 2^2}$ Ra (3 0) 2 + (2 0) 2

r₂₃ = √13

r₂₃ = 3.606 m

let's look for the job

W = U

let's calculate

W = ${9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )$

W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

W = U = k $( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )$

all charges are equal q₁ = q₂ = q₃ = q₄ = q

W = k q² $(\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )$

let's look for the distances

r₁₄ = $\sqrt{3^2 +4^2}$

r₁₄ = 5 m

r₂₄ = $\sqrt{3^2 + ( 4-2)^2}$

r₂₄ = √13 = 3.606 m

r₃₄ = $\sqrt{(3-3)^2 + (4-0)^2}$

r₃₄ = 4 m

we calculate

W = 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )$

W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

W = U = k $( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )$

in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value

W = - k q² $( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )$

we look for distances

r₁₅ = $\sqrt{ 1^2 +1^2}$ Ra (1-0) 2 + (1-0) 2

r₁₅ = √ 2 = 1.4142 m

r₂₅ = $\sqrt{ (2-1)^2 +1^2}$

r₂₅ = √2 = 1.4142 m

r₃₅ = $\sqrt{ ( 3-1)^2 +1^2}$

r₃₅ = √5 = 2.236 m

r₄₅ = $\sqrt{ (3-1)^2 + (4-1)^2}$

r₄₅ = √13 = 3.606 m

we calculate

W = - 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )$

W = - 4.93 10⁻²⁸ J

3 0

3 years ago

Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it

olga2289 [7]

<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>

6 0

4 years ago

Larry sees a group of people weeping, with frowns on their faces and their eyes turned down. Larry their expressions to understa

Salsk061 [2.6K]

Answer:

c

Explanation:

no need explanation u can trust me

8 0

3 years ago

How many outer-orbital electrons are found in an atom of:a) Na?b) P?c) Br?d) I?e) Te?f) Sr?

lys-0071 [83]

Answer:

Na=1
P=5
Br=7
I=7
Te=6
Sr=2

6 0

2 years ago