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3 years ago

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You are given two positively charged particles, of equal magnitude, separated by a distance, "d". What will happen to the force

field between the two particles when "d" is doubled? A) remain constant B) decrease by a factor of two C) decrease by a factor of four D) nothing; no force field between particles

2 answers:

lutik1710 [3]3 years ago

8 0

Explanation :

There exists a force between charged particle given by Coulomb's law.

$F=k\dfrac{q_1q_2}{d^2}$

where,

k is the electrostatic constant

$q_1,q_2$ are charged particles

d is the distance between the charges.

When d is doubled (d' = 2d) , let F' be the force.

$F'=k\dfrac{q_1q_2}{(2d)^2}$

$F'=k\dfrac{q_1q_2}{4d^2}$

$F'=\dfrac{1}{4}F$

So, the force decrease by a factor of four.

Hence, the correct option is (C) "decrease by a factor of four".

gizmo_the_mogwai [7]3 years ago

7 0

force goes as 1/d^2 ... (2d)^2 => 4d^2 ...

C) decrease by a factor of four

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an object of mass 1 g is hung from a spring and set in oscillatory motion .At t=0 the displacement is 43.75cm and the accelerati

charle [14.2K]

The spring constant is $4.0\cdot 10^{-5} N/m$

Explanation:

For an object in a simple harmonic motion, the acceleration of the object is related to the displacement by

$a=-\omega^2 x$

where

a is the acceleration

$\omega$ is the angular frequency

x is the displacement

The angular frequency is defined as

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Substituting the second equation into the first one, we get

$a=-\frac{k}{m}x$

In this problem we have

m = 1 g = 0.001 kg

And at t=0,

x = 43.75 cm

a = -1.754 cm/s

Therefore, we can re-arrange the equation above to find the spring constant:

$k=-\frac{ma}{x}=-\frac{(0.001)(-1.754)}{43.75}=4.0\cdot 10^{-5} N/m$

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3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

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3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

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3 years ago

A piece of string 2 meters and has a mass of 5g. On one end of the string hangs a 200 g mass. Find the tension of the string and

viktelen [127]

Explanation:

It is given that,

Length of the string, l = 2 m

Mass of the string, $m=5\ g=5\times 10^{-3}\ kg$

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1. The tension in the string is given by :

$T=m'g$

$T=0.2\times 9.8$

T = 1.96 N

2. Velocity of the transverse wave in the string is given by :

$v=\sqrt{\dfrac{T}{M}}$

m = M/l

$v=\sqrt{\dfrac{Tl}{m}}$

$v=\sqrt{\dfrac{1.96\times 2}{5\times 10^{-3}}}$

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sweet [91]

Answer:

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Explanation:

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