Two oppositely charged objects are separated by a small distance. The objects are then moved three times farther apart from each other. Which of the following statements best describes what happens to the electrical force between the objects?

A: The force of attraction increases
B: The force of attraction decreases
C: The force of attraction stays the same
D: The force of attraction becomes a force of repulsion

Answers

Answer 1
Force is indirectly proportional to the distance. Therefore when the distance increases the force of attraction increases and the answer would be B
Answer 2

The statement best describes what happens to the electrical force between the objects when moved three times farther apart from each other is the force of attraction decreases. The correct option is B.

What is force of attraction?

The objects when placed at a distance, they are attracted by the gravitational force of attraction.

Gravitational force of attraction is directly proportional to the product of masses and inversely proportional to the square of distance between them.

Two oppositely charged objects are separated by a small distance. The objects are then moved three times farther apart from each other.

Force is indirectly proportional to the distance. So, when the distance increases the force of attraction decreases.

Thus, the correct option is B.

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Related Questions

Leakage of liquid petroleum gas can be detected from a distance . name the process ​

Answers

Answer:

ultrasonic

Explanation:

they detect the acoustic emission created when a pressured gas expands in a low pressure area through a small orifice(the leak).

A solid aluminum sphere of radius R has moment of
inertia I about an axis through its center. What is the
moment of inertia about a central axis of a solid
aluminum sphere of radius 2R?
1. 21
2. 41
3. 87
4. 161
5. 321​

Answers

Answer:

5. 32I

Explanation:

The moment of inertia of a solid sphere about its central axis is given by

I = [tex]\frac{2}{5} MR^2[/tex]            ------------------(i)

Where;

M = mass of the sphere

R = radius of the sphere.

From the question;

Case 1: The aluminum sphere has a radius R and moment of inertia I.

This means that we can substitute these values of R and I into equation (i) and get;

I = [tex]\frac{2}{5} MR^2[/tex]       --------------(ii)

M is the mass of the aluminum sphere and is given by;

M = pV

Where;

p = density of aluminum

V = Volume of the sphere = [tex]\frac{4}{3} \pi R^3[/tex]

=> M = p([tex]\frac{4}{3} \pi R^3[/tex])              --------------------(*)

Case 2: An aluminum sphere with a radius of 2R instead.

Let the moment of inertia in this case be I' and mass be M'

Substituting R = 2R, M = M' and I = I' into equation (i) gives

I' = [tex]\frac{2}{5} M'(2R)^2[/tex]       ------------------(iii)

Where;

M' = pV'

p = density of aluminum

V' = volume of the sphere = [tex]\frac{4}{3} \pi (2R)^3[/tex]

=> M' = p([tex]\frac{4}{3} \pi (2R)^3[/tex])

Rewriting gives;

M' = p([tex]\frac{4}{3} \pi (2)^3(R)^3[/tex])

M' = p([tex]\frac{4}{3} \pi8(R)^3[/tex])

M' = 8p([tex]\frac{4}{3} \pi R^3[/tex])

From equation (*), this can be written as

M' = 8M

Now substitute all necessary values into equation (ii)

I' =  [tex]\frac{2}{5} M'(2R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(2R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(2)^2(R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(4)(R)^2[/tex]

I' =  [tex]\frac{2}{5} (32M)(R)^2[/tex]

I' =  [tex]32[\frac{2}{5}MR^2][/tex]

Comparing with equation (ii)

I' =  [tex]32[I][/tex]

Therefore, the moment of inertia about a central axis of a solid

aluminum sphere of radius 2R is 32I

A 25N[L] force acts on a 1kg object that is already travelling at 10m/s[R]. The force acts for t=2s. What is its final speed? *
A) 25m/s
B) 25m/s/s
C) 40m/s[R]
D) 40m/s[L]
show your full work please

Answers

Answer:

C) 40m/s [R]

Explanation:

Given the following data;

Force = 25 N [L]

Mass = 1 kg

Initial velocity = 0 m/s

Velocity = 10 m/s [R]

Time = 2 seconds

To find the final velocity;

First of all, we would determine the acceleration of the object.

Force = mass * acceleration

25 = 1 * acceleration

Acceleration = 25/1

Acceleration = 25 m/s²

Next, we would determine the final velocity by using the first equation of motion;

[tex] V = U + at[/tex]

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the formula, we have;

[tex] V = 0 + 25*2[/tex]

[tex] V = 0 + 50 [/tex]

V = 50 m/s

Final velocity = 50 - 10

Final velocity = 40 m/s [R]

help me with this please

Answers

Answer:

Where is the Question?

Explanation:

I am glad to h help you out.

Answer:

Sorry is there any attachments to put with this question, If please publish question properly please so that brainly answerers will know what to answer.

what was the significance of jumping a.keep the snake b.keep feet cleans c.avoid the hot water d.avoid the Bumbo stick

Answers

Answer:

D I think I’m not for sure

Explanation:

5. The disks, or pads, that exist between bones as gliding joints are made of which substance?

Answers

Answer:

cartilage

Explanation:

Answer:

Cartilage

Explanation:

Correct in AE

Under which conditions would a nucleus be least likely to undergo radioactive decay? A. When the strong nuclear forces are less than the electrostatic forces of repulsion
B. When the strong nuclear forces are greater than the electrostatic forces of repulsion
C. When the strong nuclear forces are less than the electrostatic forces of attraction
D. When the strong nuclear forces are greater than the electrostatic forces of attraction

Answers

Answer:

b is the correct answer i believe

Explanation:

the strong nuclear force encourages particles to come together so that they can nuetralize, a process the opposite of the electrostatic force of repulsion. This lets the atom stay intact because if the repulsion was more than the nuclear forces strength the atom would lose electrons and slowly decay

a brick of mass 0.8 kg is accidentally dropped from a high scaffolding. it reaches the ground with a kinetic energy of 240 J. How high is scaffolding ?(Take acceleration due to gravity g be 10 m s-¹)​

Answers

Answer:

30 m

General Formulas and Concepts:

Energy

Gravitational Potential Energy: [tex]\displaystyle U_g = mgh[/tex]

m is mass (in kg)g is gravityh is height (in m)

Kinetic Energy: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]

m is mass (in kg)v is velocity (in m/s²)

Law of Conservation of Energy

Explanation:

Step 1: Define

Identify variables

[Given] m = 0.8 kg

[Given] g = 10 m/s²

[Given] U = 240 J

[Solve] h

Step 2: Solve for h

[LCE] Substitute in variables [Gravitational Potential Energy]:                      (0.8 kg)(10 m/s²)h = 240 JMultiply:                                                                                                             (8 kg · m/s²)h = 240 JIsolate h [Cancel out units]:                                                                             h = 30 m

A bullet is shot straight up into the air from ground level. It reaches a maximum

height at h = 739 m.

Answers

Complete question:

A bullet is shot straight up into the air from ground level. It reaches a maximum height at h = 739 m. Calculate the initial velocity of the bullet.

Answer:

the initial velocity of the bullet is 120.35 m/s

Explanation:

Given;

maximum height reached by the bullet, h = 739 m

let the initial velocity of the bullet = u

At maximum height the final velocity of the bullet, v = 0

Apply the following kinematic equation to determine the initial velocity of the bullet.

v² = u² - 2gh

0 = u² - 2gh

u² = 2gh

u = √2gh

u = √(2 x 9.8 x 739)

u = 120.35 m/s

Therefore, the initial velocity of the bullet is 120.35 m/s

You sit on ice and shove a heavy box with your feet with a given force. What will you and the box share? *
A) Same acceleration
B) equal and opposite acceleration
C) the equal and opposite force
D) same force
explain please

Answers

Answer:

the correct answer is C

Explanation:

In this exercise we will analyze the situation.

When the person is on the ice, the friction coefficient is very small, if the box is in a place where there is no ice, the coefficient is different, so the friction force on each body is different.

Therefore the acceleration that each body acquires is different.

If we apply the conservation of momentum, each body moves in the opposite direction, but with different speeds.

If we use Newton's third law, the force applied to each body has the same magnitude and opposite direction, which is why the force is of the action and reaction type

Consequently the correct answer is C

you are traveling in a convertible with the top down. the car is moving at a constant velocty of 25 m/s due east along falt ground. you throw a tennis ball straight upward at a speed of 15 m/swhen the ball just leaves your hand, what is the speed of the ball as measured by an observer on the side of the road

Answers

Answer:

Teh velocity of ball with respect to the ground is 29.2 m/s.

Explanation:

Velocity of car, vc = 25 due m/s east

velocity of ball with respect to car, v(b,c) = 15 m/s due Z axis

Write the velocities in the vector form

[tex]\overrightarrow{v}_{c}=25 \widehat{i}\\\\\overrightarrow{v}_{(b,c)}=15 \widehat{k}[/tex]

The velocity of ball with respect to ground is

[tex]\overrightarrow{v}_{(b,c)}= \overrightarrow{v}_{(b,g)}-\overrightarrow{v}_{(c,g)}\\\\15 \widehat{k} = \overrightarrow{v}_{(b,g)} - 25 \widehat{i}\\\\\overrightarrow{v}_{(b,g)} = 25 \widehat{i} + 15 \widehat{k}\\\\v_{(b,g)}=\sqrt{25^2 + 15^2}=29.2 m/s[/tex]

1. Given an object that follows this time-dependent velocity function: ~v(t) = 2 m/s 2 tˆi − 3 m/s 3 t 2ˆj, and assuming the object begins at the origin at t=0s, where will the object be at t=2.0s?

2. Suppose an object of mass 2.0kg begins at rest and is acted upon by a force F(t) = 5.0N e^−0.1t . What will the object’s speed be after ten seconds?

Answers

Answer:

Explanation:

V=ds/dt

Where s =distance traveled

Given

V= 2ti - 3t²j

ds/dt=2ti - 3t²j

ds=(2ti - 3t²j)dt

Integrating

S= 2t²i/2 - 3t³j/3 + C

S=t²i - t³j + C

Since the object starts from Rest when t=0 and s(distance)=0

0=0²i - 0³j + C

C=0

Therefore

S=t²i - t³j

At t=2sec

S=(2)²i - (2)³j

S=4i - 8j

Magnitude of S(distance) = √4²+(-8)²

S= 4√5 meters

S=8.94meters.

2.mass =2kg

F(t)=5e^-0.1t

From Newton 2nd Law

F(t) = mdv/dt

5e^-0.1t= 2dv/dt

2dv = (5e^-0.1t)dt

Integrating

2V(t) = 5e^-0.1t/(-0.1) + C

2v(t) = - 50e^-0.1t + C

Since it started from rest at t=0. That is v(0)=0

2(0) = -50e^-0.1(0) + C

0 = -50 + C

C= 50

v(t) = -50e^-0.1t + 50

At t=10sec

v(10) = -50e^-0.1(10) + 50

V= -18.39 + 50

V= 31.61ms-¹.

While using your calc to evaluate -50e^-0.1(10)

Don't forget the -(minus) sign at the top of the exponential.

If you neglect it... You'll have a different answer.

Hope this helps.

Have a great day!!!

A speedometer in a car gives the car’s speed at that given moment, or the?

A. General speed
B. Instantaneous speed
C. Average speed
D. Constant speed

It’s not C or D!

Answers

Answer:

a because it is at a given moment

Explanation:

did u

A car with a mass of 1 892kg, traveling at 12m/s, was heading towards a wall. It applied 70N of force on its brakes for 6 seconds until it slows down.

Find:
1. The momentum of the car.

2. The resulting impulse of the car. ​

Answers

Answer:

MOMENTUM=205 IMPULSE = 5

Explanation:

The momentum of car is 22704 kg m/s.

The resulting impulse of car is 420 N-s.

What is momentum?

Momentum is product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.

p= mv

p= 1892*12

p= 22704 kg m/s

What is impulse?

Impulse is the motion produced by a starting force.

[tex]\Delta p = F\Delta t[/tex]

[tex]\Delta p = 70*6\\\Delta p = 420 Ns[/tex]

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cho hệ cơ học như hình vẽ hai đầu dây buộc hai vật có khối lượng tương ứng là m1=2kg và m2>m1 lấy g=10m/s sau 1s kể từ lúc bắt đầu chuyển dộng hệ vật đi được 50 cm tính m2 và sức căng của dây

Answers

xin lỗi không có sơ đồ vui lòng cho biết sơ đồ

(4.56 x 10^-13)-(1.17 x 10^-13)

Answers

3.39 x 10^-13

Please mark brainliest!

A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.240 T/s .

Requried:
If the loop has a resistance of 0.700Ω. Find the current induced in the loop.

Answers

Answer:

The induced current is 26.7 mA

Explanation:

Given;

area of the loop, A = 0.078 m²

initial magnetic field, B₁ = 3.8 T

change in the magnetic field strength, dB/dt = 0.24 T/s

The induced emf is calculated as;

[tex]emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V[/tex]

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA[/tex]

Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.

Answers

Answer:

[tex]F = 14.2 N[/tex]  

Explanation:

From the question we are told that:

Mass [tex]m=0.15kg[/tex]

Radius [tex]r=0.6[/tex]

Angular Velocity [tex]\omega=2rev/s[/tex]

                            [tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]

Generally the equation for Force applied is mathematically given by

 [tex]F =mrw2[/tex]

 [tex]F=0.15*0.6* (4*x3.14^)2[/tex]

 [tex]F = 14.2 N[/tex]    

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?

Answers

Answer:

[tex]v=6.65m/sec[/tex]

Explanation:

From the Question we are told that:

Mass [tex]m=97.6[/tex]

Coefficient of kinetic friction  [tex]\mu k=0.555[/tex]

Generally the equation for Frictional force is mathematically given by

 [tex]F=\mu mg[/tex]

 [tex]F=0.555*97.6*9.8[/tex]

 [tex]F=531.388N[/tex]

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 [tex]a_f=-\mu g[/tex]

 [tex]a_f=-0.555 *9.81[/tex]

 [tex]a_f=-54455m/sec^2[/tex]

Therefore

 [tex]v=u-at[/tex]

 [tex]v=0+5.45*1.22[/tex]

 [tex]v=6.65m/sec[/tex]

bowling ball weighing 710 N attached to the ceiling by a rope of length 3.87 mThe ball is pulled to one side and released it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.51 m/s

What is the acceleration of the bowling ball, in magnitude and direction, at this instant?

What is the tension in the rope at this instant?

Answers

Answer:

a = 5.256 m / s²  with the vertical direction upwards and   T = 1090.8 N

Explanation:

to find the acceleration we must use Newton's second law, in the lowest part of the motion

          T - W = m a

where the acceleration is centripetal

         a = v² / r

let's calculate

         a = 4.51² / 3.87

         a = 5.256 m / s²

As the acceleration is centripetal it is directed towards the center of the circle, which in the lower part coincides with the vertical direction upwards.

Let's find the tension of the rope with the first equation

        T = W + m a

         

        W = m g

let's calculate

         T = 710 + 710 5.256 / 9.8

         T = 1090.8 N

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.

Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?

Answers

Answer:

a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

b) [tex]a=30.7[/tex]

c) [tex]a=35.91[/tex]

Explanation:

From the question we are told that:

Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

Initial Length [tex]L_1=0.600m[/tex]

Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]

Final Length [tex]L_2=0.900m[/tex]

a)

Generally the rotation with the greater speed is

[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

b)

Generally the equation for centripetal acceleration at 8.16 is mathematically given by

[tex]a=\omega_1^2*L_1[/tex]

[tex]a=8.16 rev/s*0.6[/tex]

[tex]a=30.7[/tex]

c)

At 6.35 rev/s

[tex]a=6.35 rev/s*0.9[/tex]

[tex]a=35.91[/tex]

A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficients of static and kinetic friction are 0.697 and 0.371, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? If the block does not move, give 0 m/s2 as the acceleration?

Answers

Answer:

Yes it will move and a= 4.19m/s^2

Explanation:

In order for the box to move it needs to overcome the maximum static friction force

Max Static Friction = μFn(normal force)

plug in givens

Max Static friction = 31.9226

Since 36.6>31.9226, the box will move

Mass= Wieght/g which is 45.8/9.8= 4.67kg

Fnet = Fapp-Fk

= 36.6-16.9918

=19.6082

=ma

Solve for a=4.19m/s^2

show your full steps

Answers

Answer:

x = 13552.6 m

Explanation:

This is a projectile throwing exercise

Let's start by looking for the time it takes for the pump to reach the ground y = 0

         y = y₀ + v_{oy} t - ½ g t²

as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t²

         t = [tex]\sqrt{2y_o/g}[/tex]

         t = [tex]\sqrt{2 \ 10000/9.8}[/tex]

         t = 45.175 s

with this time we can find the distance it travels horizontally

         x = v₀ₓ t

         x = 300 45,175

         x = 13552.6 m

the bomb must be dropped at this distance before hitting the target

Planets closer to a star will have what type of average temperature

Answers

Answer:

Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018

Question 1 of 25
Which equation is an example of a synthesis reaction?
A. HNO3 + KOH → KCI + H20
B. 2Li+ CaCl2 - 2LiCl + Ca
O C. S+ 02 - S02
7
O D. CH4 + 202 - 2H2O + CO2

Answers

Answer:

C. S + 02 → S02

Explanation:

A synthesis or combination reaction is that reaction involving two elements as reactants to form a single compound as product.

In the reaction given below;

S + 02 → S02

Sulphur and oxygen are elemental substances that combine to synthesize sulfur IV oxide (SO2), and hence it is an example of synthesis reaction.

While taking a psychology exam, Lori got up and shut the classroom door due to outside noise that affected her concentration. Which theory explains Lori’s behavior?

Group of answer choices

Drive-reduction theory

Expectancy theory

Instinct theory

Arousal theory

Answers

Answer:

drive reduction theory

Explanation:

I would say that because of all the cars beeping and making A Lot of cachos on the street so that will definitely affect her taking her exam

Theory explains Lori’s behavior is b) Drive-reduction theory

What is Drive-reduction theory?

It is based on the idea that  the primary motivation behind all human behavior is to reduce drives . A drive is a state of  discomfort which is triggered by a person's physiology or biological need .

so , she removed discomfort that can happen by the noise outside the classroom in order to maintain student's concentration

correct answer is b) Drive-reduction theory

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Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of charge from one sphere to the other? Answer in Joules.

Answers

Answer:

Use 2>60n to get 19>089

Explanation:

New alleles arising from mutations in a population will​

Answers

Increase in frequency over time until they reach fixation, replacing the ancestral allele in the population

High speed stroboscopic photographs show that the head of a 183 g golf club is traveling at 58.6 m/s just before it strikes a 46.6 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.

Answers

Answer:

The speed of the golf ball just after the impact is 73.04 m/s.

Explanation:

Given that,

The mass of golf club, m₁ = 183 g = 0.183 kg

The mass of golf ball, m₂ = 46.6 g = 0.0466 kg

The initial speed of golf club, u₁ = 58.6 m/s

The initial speed of a golf ball, u₂ = 0

The final speeds of club, v₁ = 40 m/s

We need to find the speed of the golf ball just after impact. Using the conservation of momentum to find it.

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.183 (58.6)-0.183(40)}{0.0466 }\\\\=73.04\ m/s[/tex]

So, the speed of the golf ball just after the impact is 73.04 m/s.

A 2.80 kg mass is dropped from a
height of 4.50 m. Find its potential
energy (PE) when it is 3.00 m
above the ground.

Answers

Answer:

PE = 82.32J

Explanation:

PE = m*g*h

PE = 2.80kg*9.8m/s²*3m

PE = 82.32J

Mark Brainliest please

Answer : 41.2 J

Explanation

The energy that the ball had instantaneously prior to being dropped was Potential Energy (EP) only and gravitational potential energy is calculated as Ep = mgh = (2.80)(9.81)(4.50) = 123.606 Joules.

Because no outside energy was added and no work was done, the energy instantaneously prior to being dropped is equal to the energy when the mass is 3 m above the ground only at that point, the energy consists of two parts, gravitational Potential Energy (EP) and Kinetic Energy (EK). So, considering the point just prior to the drop as point 1 and the point 3 m above the ground as point 2, we can say:
EP1 = EP2 + EK2
so EK2 = EP1 - EP2
and EP2 = (2.80)(9.81)(3.00) = 82.404
So EK2 = 123.606 - 82.404 = 41.2 Joules
Other Questions
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