Physics, 25.03.2020 16:58 1031kylepoe03

Two particles, with identical positive charges and a separation of
2.60×10−2m
are released from rest. Immediately after the release, particle 1 has an acceleration
→
a
1
whose magnitude is
4.60×103m/s2
while particle 2 has an acceleration
→
a
2
whose magnitude is
8.50×103m/s2
Particle 1 has a mass of
6.00×10−6kg
Find (a) the charge on each particle and (b) the mass of particle 2.

Answers: 1

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Answer from: moynihansavanah003

(a)Therefore the charge of each particle is $4.56\times 10^{-4} \ C$ .

(b)Therefore the mass of particle 2 is 3.25×10⁻⁶ Kg.

Explanation:

Given that, two positive charges particles are equal charge and the distance between them is 2.60×10⁻² m²

The acceleration of the first particle is 4.60×10³ m/s².

The acceleration of the second particle is 8.50×10³ m/s².

The mass of the first particle is 6.00×10⁻⁶kg.

The coulombs Law:

$F=\frac{kq_1q_2}{r^2}$

F= Force between the charges.

$q_1 \ and \q_2$ = The charges

r= Distance between the changes.

Here q_1 =q_2=q ,r = 2.60×10⁻² m $k=9\times 10^9$ Nm²/C²

Therefore

$F=9\times 10^9.\frac{q^2}{(2.60\times 10^{-2})^2}$

=1.33×10¹³ ×q² N.

Again we know that,

Force = mass× acceleration

For the first particle m₁=6.00×10⁻⁶kg, a₁= 4.60×10³ m/s² and F =1.33×10¹³ ×q² N

∴1.33×10¹³ ×q² =6.00×10⁻⁶×4.60×10³

$\Rightarrow q^2=\frac{6.00\times 10^{-6}\times 4.60\times 10^3}{1.33\times 10^{13}}$

$\Rightarrow q^2= 20.75\times 10^{-16}$

$\Rightarrow q=4.56\times 10^{-4}$ C

Therefore the charge of each particle is $4.56\times 10^{-4} \ C$ .

(b)

To find the mass of the second particle we use the following formula,

Force = mass× acceleration

Here F=m₁a₁, mass = m₂, acceleration= a₂=8.50×10³ m/s².

m₁a₁= m₂a₂

⇒(6.00×10⁻⁶)×( 4.60×10³)= m₂×(8.50×10³)

$\Rightarrow m_2=\frac {(6.00\times10^{-6})\times( 4.60\times10^{3})}{(8.50\times10^3)}$

=3.25×10⁻⁶ Kg.

Therefore the mass of particle 2 is 3.25×10⁻⁶ Kg.

Answer from: Quest

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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Two particles, with identical positive charges and a separation of
2.60×10−2m
are releas...

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