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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Answers

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

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

490 n

f=125n a=2.5m/s2

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