Two particles with charges q1 = q2 = 7.00 nC are separated by a distance of 1.00 m. Suppose that q1 is at the point x = -0.500 m
a) Make a diagram of the situation.
b) At what point on the line connecting the two charges, is the total electric field produced by both charges equal to zero?
c) Suppose that a third electric charge q3 = 1.00 nC and mass m = 1 mg is placed at the point where the electric field is equal to zero and is given a small push. Determine the approximate value of the angular frequency of oscillation of q3 around this point, assuming small displacements around the equilibrium point .
Hint: Use the binomial formula appropriately to make this approximation: (1 + x) n ≈ 1 + nx.
Answers
The definition of electric field, Coulomb's law and Newton's second law allow to find the results for the questions about the electric field and the oscillation of the test charge are:
a) In the attachment you have the scheme of the system
b) The point where the electric field is zero is: x = 0 m
c) The angular velocity of the test charge is w = 0.5 rad / s
Given parameters
Value of the charges q₁ = q₂ = 7.0 10⁻⁹ C Separation of charges d = 1.0 m Charge position q₁ is: x₁ = -0.5 m The value of the charge 3 is q₃ = 1.00 10⁻⁹CTo find
a) system scheme
b) point where the field is zero
c) The angular velocity of q₃
The electric field is the field produced by a distribution of electric charges at a point in space, it is a vector quantity, for the special case that the charges are in one dimension the sum of the field produced by a series of charges is reduced to the sum algebraic. The field produced by point charges is:
E = [tex]k \sum \frac{q_i}{r_i^2}[/tex]
Where E is the electric field, k is the Coulomb constant, q the value of the charge and r the distances from the charge to the point of interest.
a) In the attachment we can see a diagram of the system, the reference system, the location of the charges and the electric field vectors are shown.
b) From the diagram we see that each field is in the opposite direction, therefore the total field is:
[tex]E_{total} = E_1 - E_2 \\E_{total}= k \frac{q_1}{(-0.5- x)^2} - k \frac{q2}{(0.5-x)^2 }\\\\E_{total = 0[/tex]
[tex]\frac{q}{(0.5+x)^2 } = \frac{q}{(0.5-x)^2}[/tex]
The value of the two charges is the same, therefore the value of position x is:
x = 0 m
The point where the total electric field becomes zero is at the origin of the coordinate system.
c) We look for the value of the force on the test charge introduced q₃, for this using the Coulomb equation that establishes that the force is proportional to the product of the electric charges and is inversely proportional to the square of the distance
[tex]F_{13} = k \frac{q_1 q_3x}{(x+0.5)^2}\\F_{23} = k \frac{q_2 q_3}{(x-0.5)^2 }[/tex]
Newton's second law gives the relationship between the net force and the product of mass and acceleration.
[tex]\sum F = m \a \\k q q_3 [ \frac{1}{(x+0.5)^2} - \frac{1}{(x-0.5)2 } }] = m \frac{d^2x}{dt^2}[/tex]
if we use the binomial development
(0.5 + x)⁻² = 0.5 +2 x
(0.5 -x)⁻² = 0.5 -2x
we substitute
k q q₃ [-4x] = [tex]m \frac{d^2x}{dt^2}[/tex]
[tex]-(\frac{4k \ q \ q_3}{m}) \ x = m \frac{d^2x}{dt^2}[/tex]
In oscillatory motion the general equation of motion is
[tex]- w^2 \ x = \frac{d^2x}{dt^2}[/tex]
Equating the two expressions the angular velocity is
[tex]w^2 = \frac{4k \ q \q_3}{m}[/tex]
Let's calculate.
w²= [tex]\frac{4 \ 9 \ 10^9 \ 7 \ 10^{-9} \ 1 \ 10^{-9}}{1 \ 10^{-6}}[/tex] a 4 9 10⁹ 7 10-9 1 10-9 / 1 10-6
w =[tex]\sqrt{0.252}[/tex]
w = 0.5 rad / s
In conclusion, using the definition of electric field, Coulomb's law and Newton's second law we can find the results for the questions about the electric field and the oscillation of the test charge are:
a) In the attachment you have the scheme of the system.
b) The point where the electric field is zero is: x = 0 m
c) The angular velocity of the test charge is w = 0.5 rad / s
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Related Questions
1
2) Mighty Mouse can lift seven mammoths with a total mass of 525 kg. Using the acceleration of
gravity, find the total WEIGHT that Mighty Mouse is lifting. (answer: 5250 N)
Answers
Answer:
5250N
Explanation:
Weight= Mass × acceleration due to gravity
= 525kg × 10m/s^2
= 5250N
1.25 is closer to 1.04 or not ?
plz heelp plz
Answers
Answer:
No, it is closer to 1.30
Explanation:
A meter stick is attached to one end of a rigid rod with negligible mass of length l = 0.302 m. The other end of the light rod is suspended from a pivot point, as shown in the figure below. The entire system is pulled to a small angle and released from rest. It then begins to oscillate. A meter stick hung from a rod of length l. The rod is attached to the ceiling. The rod and meter stick extend downward in a straight line making a small angle with the vertical. (a) What is the period of oscillation of the system (in s)? (Round your answer to at least three decimal places.)
Answers
The period of oscillation of the system nearest to three decimal places
= 1.092 seconds
The period of an oscillation occurring in a system is the time taken to complete one cycle.
The formula that is used to calculate the period of oscillation (T) is
= 2π√[tex]\frac{l}{g}[/tex]
But,
π = 3.14159 (constant)
g= 10m/s² (acceleration due to gravity)
l = 0.302 m
Therefore T = 2 × 3.14159 × √[tex]\frac{0.302}{10}[/tex]
= 6.28318 x √0.0302
= 6.28318 x 0.17378
= 1.09189s
= 1.092 seconds ( to the nearest three decimal places)
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Acceleration of a car that speeds from 4.3 m/s to 12.9 m/s in 2 seconds
Answers
Explanation:
let v1 = 4.3 m/s
v2 = 12.9 m/s
t = 2 seconds
v2 = v1 + at
12.9 = 4.3 + a×2
2a = 12.9 - 4.3 = 8.6
a = 8.6/2
a = 4.3 m/s^2
Option B.
Consider a setup in which two springs are attached to a mass in parallel.
Convince yourself that in this setup, the compression of each spring must be the same. Using
this fact, derive the effective spring constant for springs in parallel
This is asking, "ll1 replace the two springs by a single imaginary spring, what would its spring
constant be such that the force stays the same?" Your answer should only depend on k, and k
Answers
Answer:
it would be...
Explanation:
A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during the first 2 minutes of flight, the main engines operate for a total of 8.5 minutes after the launch. Once the SRBs are released, the main engines alone accelerate the rocket from about 1341 m/s to 7600 m/s.
What is the acceleration of the SRB and main engine during the first 2.0 minutes of flight?
A. 52 m/s2
B. 13 m/s2
C. 9.8 m/s2
D. 11 m/s2
Answers
A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:
From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:
The speed increases from 0 m/s to 1341 m/s.The time elpased is 2.0 min.1 min = 60 s.The acceleration of the ship during the first 2.0 minutes is:
[tex]a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s} = 11 m/s^{2}[/tex]
A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
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What is non examples of enlarged
Answers
reduction
deflation
these are examples of the opposite of enlarged to make something smaller is really the key thing here
Which of the following scenarios would have the MOST friction force between the object and the surface?
A. a 7 kilogram bicycle with rubber tires on concrete
B. a 7 kilogram bicycle with rubber tires on ice
C. a 0.25 kilogram toy car with rubber tires on ice
D. a 0.25 kilogram toy car with rubber tires on concrete
Answers
Explanation: mass and friction are proportional if the mass increases the friction increases
Which force, in real life, will have the least effect on a bowling ball rolling down a lane toward bowling pins?
A) magnetism
B) air resistance
C) gravity
D) friction
Answers
Answer:
Its Friction
Explanation:
the pins are not floating and they are not a magnet and not involved with air
The force, in real life, that will have the least effect on a bowling ball rolling down a lane toward bowling pins is magnetism. The correct option is A.
What is magnetism?Magnetism is basically the force which indeed magnets exert when they attract or even repel one another. The movement of electric charges resulting in magnetism.
Every substance is composed of tiny units referred to as atoms. Each atom contains electrons, which are charged particles.
To increase stability, the pins themselves have a low center of gravity. Because they are spherical in shape, they can roll and strike other pins in a variety of directions.
The force acting on the bowling ball is friction and air resistance. The friction force is equal to the friction coefficient multiplied by the normal force, and thus mass times acceleration.
Thus, the correct option is A.
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#SPJ2
When an elastic object is changed from its original shape:
A:Energy is released
B:Work is done
C:It is ruined
D:It makes a twanging sound
Answers
Answer:
deformation : elastic deformation is reversed when the force is removed. inelastic deformation is not fully reversed when the force is removed – there is a permanent change in shape.
Explanation:
tysm
Which of Newton’s Laws of motion help explain the following scenario?
A paddle wheel boat pushes on the water and the water pushes back causing the boat to move
Answers
Answer:
Newton #3
Explanation:
For every action there is an equal and opposite reaction.
Market researchers were interested in the relationship between the number of pieces in a brick-building set and the
cost of the set. Information was collected from a survey and was used to obtain the regression equation ý =
0.08x + 1.20, where x represents the number of pieces in a set and ŷ is the predicted price in dollars) of a set.
What is the predicted price of a set that has 500 pieces?
$40
$41.20
$600
$6,235
Answers
Then do 40 + 1.2 = 41.20
Answer B
 what is the difference between repelling and attracting
Answers
Answer:
Attracting means pulling toward you and repelling means pushing away
Explanation:
Answer: Repelling is when something will not connect with another object. The force will cause a repel between the two objects. Attracting is when something is attracted or being pulled to another object.
Explanation: Hope this helps!
A girl of mass m1=60.0 kilograms springs from a trampoline with an initial upward velocity of vi=8.00 meters per second. At height h=2.00 meters above the trampoline, the girl grabs a box of mass m2=15.0 kilograms. (Figure 1)
For this problem, use g=9.80 meters per second per second for the magnitude of the acceleration due to gravity.
What is the speed vbefore of the girl immediately before she grabs the box?
Express your answer numerically in meters per second.
What is the speed vafter of the girl immediately after she grabs the box?
Express your answer numerically in meters per second.
What is the maximum height hmax that the girl (with box) reaches? Measure hmax with respect to the top of the trampoline.
Answers
The conservation of momentum and energy allows to shorten the results for the movement of the girl on the trampoline holding the box are:
a) the girl's speed is v = 4.98 m / s
b) The speed of the girl + box system is: v_f = 0.996 m / s
c) the maximum height is: y = 2.05 m
Kinematics studies the movement of bodies, looking for relationships between the position, velocity and acceleration of bodies.
The momentum is defined by the product of mass and the velocity, when a system is isolated the momentum is conserved.
The mechanical energy is the sum of the kinetic energy plus the potential energies, when there is no friction in the system the mechanical energy is conserved.
Let's solve this exercise in parts:
a) Let's use kinematics to find the speed of the girl before she grabs the box
v² = v₀² - 2 g y₁
v² = 8² - 2 9.8 2.00
v = R 24.8 = 4.98 m / s
b) Let's use momentum conservation for when the speed of the girl and the box together. Let's write the moment in two moments.
Initial instant. Just before you grab the box.
p₀ = M v + 0
Final moment. Right after taking the box
[tex]p_f[/tex] = (m + M) [tex]v_f[/tex]
In system this form by the girl and the box therefore it is an isolated system and the momentum is conserved.
[tex]p_o = p_f[/tex]
mv = (m + M) [tex]v_f[/tex]
[tex]v_f = \frac{m}{m+M} \ v[/tex]
Let's calculate
[tex]v_f = \frac{15}{15+ 60} \ 4.98[/tex]
[tex]v_f[/tex] = 0.996 m / s
c) Now we use conservation of energy after the girl has the box.
Starting point. When the girl has the box
Em₀ = K + U
Em₀ = ½ (m + M) v² + (m + M) g y₁
Final point. At the highest point of the trajectory
[tex]Em_f[/tex] = U
[tex]Em_f[/tex] = (m + M) g y₁
As there is no friction, the energy is conserved.
[tex]Em_o = Em_f[/tex]
½ (m + M) v² + (m + M) g y₁ = (m + M) g y
y = [tex]\frac{v^2}{2g} + y_1[/tex]
Let's calculate
y = [tex]\frac{0.996^2}{2 \ 9.8} + 2.0[/tex]
y = 2.05 m
In conclusion using the conservation of momentum and energy we can shorten the results for the movement of the girl on the trampoline holding the box are:
a) the girl's speed is v = 4.98 m / s
b) The speed of the girl + box system is: v_f = 0.996 m / s
c) the maximum height is: y = 2.05 m
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Answer:
Vbefore = 4.98 m/s
Vafter = 3.98 m/s
Hmax = 2.81 m
Alex is x years old . June is 7years older than Alex . their total combined ages is 29 years . find June,s age . show all work algebraically
Answers
Answer:
18 yearsExplanation:
Given,
Let Alex be = x years
Then June will be = (7 + x) years
We know that,
Their total combined age is 29 years
Therefore,
By the problem,
=> x + (7 + x) = 29
=> 2x = 29 - 7
=> 2x = 22
=> x = 22 ÷ 2
=> x = 11
So,
Required age of June is = (7 + x) years
= (7 + 11) years
= 18 years (Ans)
Colette launches an air rocket in the upward, positive direction. It launches with an initial velocity of 25.5 m/s. It accelerates in the downward, negative direction at a rate of 9.81 m/s2. After 3.5 seconds, what is the magnitude of the rocket's displacement?
Answers
Answer:
Give me some hint please
A proton is released in a uniform electric field and it experiences an electric force of 2.18x10^-14N toward south .What are the magnitude and direction of the electric field
Answers
Answer:
F = E q
E = 2.18E-14 / 1.6E-19 = 1.36E5 N/C
The direction of the field will be the same as force on the proton - southerly
For a horizontally projectile object if the initial velocity doubled, the Vertical displacement (angle y) will be :
Answers
Answer:
Unchanged
Explanation:
horizontal projection always has initial vertical velocity equal to zero.
gravity works pretty much the same at any point on earth and will accelerate any falling object at the same rate traveling the same vertical distance in the same time. Neglecting air resistance of course.
By the way, Δy is read "delta y" not "angle y" which would be "∠ y"
delta represents a change in something, in this case the y coordinate.
If you do 72 J of work in 1.2 seconds, how much power is produced
Answers
Answer:
Explanation:
Power is the rate of doing work
P = 72 / 1.2 = 60 Watts
A bicyclist travels 30 km in 2 hours east. Calculate its velocity?
Answers
Answer:
Its velocity would be
15 km per hour
or
0.00416667 per second
An object moves 45m due East and then 15m due West, the total distance is
Answers
Answer:30m east
Explanation: 45m-15m=30m, which cancels out the one to the west. All that is left is 30m east.
Is this a test/quiz btw?
Before the collision, the toy train is travelling at 0.5 m/s. The train and the stationary
truck both have a mass of 75 g.
Calculate the momentum of the toy train before the collision?
Answers
Answer:
Explanation:
p = mv = 0.075(0.5) = 0.0375 kg•m/s
what is the approximate distance from the surface of the earth center 2900km 700km 50000km 6400km
Answers
Answer:
6400km is the closest
Explanation:
In a car crash, how are force, mass, acceleration, and velocity related?
Answers
Answer:
the answer is simple but click on the brainiest and thank me.
Explanation:
F=force, m=mass, a=acceleration, v=velocity
v=displcement/time
a= v/t
F= m×a
F= m× (v/t)
A driver entering the outskirts of a city takes her foot off the accelerator so that the car slows down from 90 km/h to 50 km/h in 10.0s. Find the car’s average acceleration
Answers
Answer:
Explanation:
a = (vf - vi) / t
a = (50 - 90) / 10.0
a = -4 km/h/s(1000 m/km / 3600 s/h)
a = - 1.11 m/s²
What on earth is equal to 9.8m/s/s
Answers
Answer:
Acceleration due to gravity
A Scooter has a mass of 250 kg. A constant force is exerted on it for 6.0 s. During the time the force is exerted, the scooter increases its speed from 6.00 m/s to 280 m/s. What is the magnitude of the force exerted on the scooter
Answers
Answer:
917 N
Explanation:
917 N, this is your answer!!
Glad to help.
A 15 g bullet is fired at 630 m/s into a 4.9 kg block that sits at the edge of a 75-cm-high table. The bullet embeds itself in the block and carries it off the table.
How far from the point directly below the table's edge does the block land?
Answers
The horizontal distance traveled below the table's edge is 29.4 cm.
The given parameters;
mass of the bullet, = 15 g = 0.015 kgspeed of the bullet, = 630 m/smass of the block, = 4.9 kgThe final velocity of the bullet-block system is calculated as follows;
[tex]m_1 u_1 + m_2u_1 = v(m_1 + m_2)\\\\0.015(630) + 4.9(0) = v(0.015 + 4.9)\\\\9.45 = 4.915 v\\\\v = \frac{9.45}{4.915} \\\\v = 1.923 \ m/s[/tex]
The time for the bullet-block system to reach the ground from the table is calculated as follows;
[tex]h = v_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\2h = gt^2\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.75}{9.8} }\\\\t = 0.153 \ s[/tex]
The horizontal distance traveled below the table's edge is calculated as follows;
[tex]X = vt\\\\X = 1.923 \times 0.153\\\\X = 0.294 \ m\\\\X = 29.4 \ cm[/tex]
Thus, the horizontal distance traveled below the table's edge is 29.4 cm.
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50 Points Help Asap !!
Briefly describe V-T graph For every point please if its increasing if its flat or if its positive or negative help fast
Answers
#A
It's flat as no change can be seen on the graph of A part#B
Its increasing as the line goes up in B part.#C.
It's flat as no change can be seen on the graph of C part#D
It's decreasing as the graph goes down in D part.#E
It's decreasing as the graph goes down in E part#F
It's decreasing as the graph goes down in F part.#G
It's increasing as the graph goes up in G part.Which scientist is credited with having the greatest contribution to early microscopy and was the first to observe and describe single-celled organisms?
Answers
Answer:
Antonie van Leeuwenhoek
Explanation:
