Why does an iceberg have more HEAT than a cup of coffee?

Answers

Answer 1

Answer:

The hot coffee has a higher temperature, but not a greater internal energy. Although the iceberg has less internal energy per mass, its enormously greater mass gives it a greater total energy than that in the small cup of coffee.

Explanation:

Answer 2

Answer:

The hot coffee has a higher temperature, but not a greater internal energy. Although the iceberg has less internal energy per mass, its enormously greater mass gives it a greater total energy than that in the small cup of coffee.


Related Questions

a car travels at the speed of 117km/h.How far will the car travels in 50 minutes​

Answers

Answer:

117/60*50

Explanation:

Trust me bc I'm smart

117x5/6 which is 97.5km

EXPLAINATION: if it goes 117 in 60 minutes n we wanna see how far it got in 50 minutes which is 5/6 of of 60 minutes so u multiply the rate (117km) by 5/6

1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock

Answers

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the  water, V, is given as follows;

[tex]V_T[/tex] = [tex]V_r[/tex] + V

[tex]V_T[/tex] = A × h₂

∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³

The total volume, [tex]V_T[/tex] = 120 cm³

The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V

∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, [tex]V_r[/tex] = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 6.9 s later. How high is the cliff?

Answers

This is provided it starts from rest:
t=6.9 u=0 a=-9.8 (acceleration due to gravity) s=?

S=ut+0.5at^2
S=0+0.5(-9.8)(6.9^2)
S=(-4.9)(47.61)
s=-233.289m (minus is because it’s below starting point)
s=233.289metres

How long would it take a machine to do 5.000
joules of work if the power rating of the machine
is 100 watts?
A. 5,000 sec
B.
50 sec
C.
10 sec
D. 0.2 sec




Answer : B

Answers

Answer:

a

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

The given machine will take 50 s to complete the work. the power is the rate of performing work.

What is power?

It can be defined as the rate of performing work. It can also be written as the amount of work divided by the time it takes to complete the work.

[tex]p = \dfrac wt[/tex]

So

[tex]t = \dfrac w p[/tex]

Where,

[tex]p[/tex] - power = 100 watt  = 100 J/s

[tex]t[/tex] - time = ?

[tex]w[/tex]- work = 5000 J

Put the values in the formula,

[tex]t = \dfrac{ 5000 \rm \ J} {100 \rm \ J/s}\\\\t = 50 \rm \ s[/tex]

Therefore, the given machine will take 50 s to complete the work.

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A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.

Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Answers

Answer:

x = 0.056 m

ΔKE = 0.489 J

Explanation:

Given that

Angle, θ = 38°

Length, L = 1.7 m

Mass, m = 0.09 kg

Spring constant, K = 590 N/m

If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE

This is mathematically written as

1/2kx² = mgH

The height, H we can get by using the relation

H = L.Sinθ

H = 1.7 * Sin 38

H = 1.7 * 0.6157

H = 1.047 m

Next, we use the Work-Energy theorem

1/2kx² = mgH

1/2 * 590 * x² = 0.09 * 9.8 * 1.047

295 * x² = 0.9234

x² = 0.9235 / 295

x² = 0.00313

x = √0.00313

x = 0.056 m

If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8

Then we use the formula

ΔKE = mg(H - H1)

ΔKE = mg(xsinθ - x2.sinθ)

Where, x = 1.7 , x2 = 0.8

ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)

ΔKE = 0.882(1.047 - 0.493)

ΔKE = 0.882 * 0.554

ΔKE = 0.489 J

A window air conditioner that consumes 2 kW of electricity when running and has a coefficient of performance of 3 is placed in the middle of a room, and is plugged in. The rate of cooling or heating this air conditioner will provide to the air in the room when running is:

Answers

Answer: Q= QH - QL

= W + QL - QL

= W

= 1 kj/8

Explanation:

Since the net rate is positive the room will be heated.

Hello everyone I have a question for you today. So you know we have discovered black holes and have theorized about white holes so let me ask you without you going to the internet and looking it up what would happen if a white hole and a black hole collided?

Answers

Answer:

White hole is an impossible object in universe. ... This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.

Section of hollow pipe and a solid cylinder have the same radius, mass, and length. They
both rotate about their long central axes with the same angular speed. Which object has the
higher moment of inertia? (a) the hollow pipe (b) the solid cylinder (c) they have the same
rotational kinetic energy (d) impossible to determine​

Answers

Answer:

Option B (The hollow pipe has the higher angular momentum)

Explanation:

Angular momentum = [tex]I*w[/tex]

I = Moment of inertia

w = Angular velocity

Hollow pipe can be considered as hollow cylinder

Moment of Inertia of Hollow cylinder = [tex]Mr^2[/tex]

Moment of Inertia of solid cylinder= [tex]\frac{Mr^2}{2}[/tex]

Clearly Moment of inertia of hollow pipe is greater which states that Angular momentum of it will be greater.

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Sound travels at a speed of 330 meters/second. If Denise hears a police siren 150 meters away, approximately how long did it take for the siren sound to
travel from the police vehicle to her?

Answers

Answer:

It went about 2 meters away

Explanation:

What is used to measure the radiation?

Answers

Answer:

geiger counter with geiger mueller

A Geiger counter is an instrument used for detecting and measuring ionizing radiation.

Hope it helps you
Thank you!

The tendency for an object to remain at rest in continue in motion is called:

Inertia
Motion
Gravity
Force

Answers

Answer:

A Inertia

Explanation:

A: inertia


Explanation: there isn’t one I just need to put 20 words

We can not hear Infrasound. Why is that?

Answers

Answer :

Last choice

Answer:

its frequency is too low

2. A test reveals that 150 J of work is required to lift an object 3 m at a
constant speed. What is the weight of the object?
0 50 N
25N
55N
O 75N

Answers

Answer:

50N

Explanation:

W=Fd

150=F(3)

50N=F

Name one similarity and one difference between a set and a bump in volleyball??

Answers

Bump: Bumping the ball means a player uses their forearms to pass the ball to a teammate or to hit the ball back over the net to the other team. Set: Setting the ball means a player positions the ball in a way that lets a teammate spike it over the net.

One similarity is the use of physical body whereas one difference is that one is exercise and the other is a sport.

One similarity and one difference

One similarity between a set and a bump in volleyball is the movement and use of legs and hands.

Whereas one difference between a set and a bump in volleyball is that completing several reps of a particular exercise in a row is called a set while on the other hand,  the basic pass in volleyball is known as bump.

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PLEASE CLICK ON THIS IMAGE I NEED HELP

Answers

Answer:

Explanation:

you can say the law of superpoition can tell us that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock or fossil in the rock is older if it is farther down in the rock layers. hope it helps

what determines the magnification of an imagev
measure:what the current values of ​

Answers

Answer:

The magnification of an image is equal to the ratio of the image height to the object height.

A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.64 m and a mass of 0.48 kg. The rod has a length of 1.78 m and a mass of 0.50 kg. The rod is placed on a fulcrum (pivot) at X = 0.34 m from the left end of the rod.
(a) Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum. kg m2
(b) Calculate the torque about the fulcrum, using CCW as positive. N.m
(c) Calculate the angular acceleration of the contraption, using CCW as positive. rad/s2
(d) Calculate the linear acceleration of the right end of the rod, using up as positive. m/s2

Answers

The image of this hollow sphere and uniform rod is missing, so i have attached it.

Answer:

A) J = 0.7443 kg•m²

B) T = 1.9169 N•m CCW

C) α = 2.5754 rad/s²

D) a = 3.966 m/s²

Explanation:

A) The moment of inertia J of the contraption around the fulcrum is given by the formula;

J = Jℓ + Jr

Let's calculate Jℓ

Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)

Jℓ = 0.4647 kg•m²

Now, let's Calculate Jr

Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50

Jr = 0.2796 kg•m²

Thus;

J = 0.4647 + 0.2796

J = 0.7443 kg•m²

(b) Using CCW as positive, Torque in Nm is calculated as;

T = Tℓ - Tr

Let's calculate Tℓ

Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81

Tℓ = 4.7739 N•m CCW

Now, let's Calculate Tr;

Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

Tr = 2.857 N•m CW

Thus;

T = 4.7739 - 2.857

T = 1.9169 N•m CCW

(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;

α = T/J

α = 1.9169/0.7443

α = 2.5754 rad/s²

(d) The linear acceleration a of the right end of the rod, using up as positive is given by;

a = α*(1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

A) the moment of inertia of the contraption is 0.7443 kgm²

B) The torque about the fulcrum is 1.9169 Nm

C) Angular acceleration of the contraption is 2.5754 rad/s²

D) The linear acceleration of the contraption is 3.966 m/s²

Moment of inertia:

(A) The moment of inertia I of the contraption around the fulcrum is given by :

[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]

I = 0.4647 + 0.2796

I = 0.7443 kgm²

(B) Using CCW as positive, Torque in Nm is given by;

T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

T = 4.7739 - 2.857

T = 1.9169 Nm

(C) The angular acceleration (α) of the contraption is given by:

α = T/I

since, torque is defined as T = Iα

α = 1.9169/0.7443

α = 2.5754 rad/s²

(D) The linear acceleration (a) of the right end of the rod

a = αr

where r is the distance from the pivot

a = α × (1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

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A runner is moving at a speed of 20 m/s. How much distance would they cover in 10 seconds?

Answers

Answer:

200 meters

Explanation:

20 x 10 = 200

200meter down is the correct answer

An electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 5.69 x 106 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

Answers

Answer:

magnitude is 1382.59 N/C

Explanation:

Given the data in the question;

The time taken is;

t = x / v

we substitute;

t = ( 2 × 10⁻²) / ( 5.69  × 10⁶ )

t = 3.5149 × 10⁻⁹ s

next, the acceleration is;

a = 2y/t² = [2( 0.150 × 10⁻²)] / [ ( 3.5149 × 10⁻⁹ )² ]

a =  2.42826 × 10¹⁴ m/s²

now, the electric field is;

E = ma / q

we know that;

mass of electron m = 9.11 × 10⁻³¹ kg,

charge of electron q = 1.60 × 10⁻¹⁹ coulomb

we substitute

E = ( 9.11 × 10⁻³¹ )(2.42826 × 10¹⁴) / 1.60 × 10⁻¹⁹

E = 2.21214 × 10⁻¹⁶  / 1.60 × 10⁻¹⁹

E = 1.3826 × 10²¹

E = 1382.59 N/C

Therefore, magnitude is 1382.59 N/C  

A block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp

Answers

Answer:

Both.

Explanation:

Given that a block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp ?

Since both of them have the same mass and the same initial velocity, then, they will both have the same kinetic energy.

That is,

K.E = 1/2mv^2

Friction is a force that opposes motion. And since the frictional force is zero,

Both of them will accelerate from Newton's law.

F = ma

We can therefore conclude that both of them will make it further up the ramp.

someone please help I can mark brainless

Answers

The answer here is B I hope that helps you if not I’m really sorry

A disk drive plugged into a 120V outlet operates on a voltage of 12V . The transformer that powers the disk drive has 125 turns on its primary coil.
Part A) Should the number of turns on the secondary coil be greater than or less than 125? greater than or less than
Part B) Explain
Part C) Find the number of turns on the secondary coil.

Answers

Answer:

(A-) Number of turns in secondary coil will be less than 125.

(B-) EXPLAIN- The voltage is to be lowered. The voltage will be reduced from 120 to 9 volts. As a result, a step down transformer will be used, indicating that the secondary coil's number of turns would be less than 125.

(C-) The number of turns on the secondary coil is 12.5

Explanation:

STEP DOWN TRANSFORMER - A step-down transformer transforms high voltage (HV) and low current from the transformer's primary side to low voltage (LV) and high current on the secondary side. A phase down transformer is the opposite of this.

In electrical networks and transmission lines, a step-down transformer can be used in a number of ways.

Step down transformer formula-

                  [tex]V_s=\frac{N_s\times V_p}{N_p}[/tex]

We are given with the -

Primary voltage [tex]V_p[/tex] = 120V

Secondary voltage [tex]Vs=12V[/tex]

Number of primary turns [tex]N_p=125[/tex]

To calculate the number of turns on the secondary coil -

       Since , the transformer ratio is given as -

                   [tex]\frac{N_s}{N_p} =\frac{V_s}{V_p}[/tex]

So, the number of turns on the secondary coil is - [tex]N_s=N_p\times\frac{V_p}{V_s}[/tex]

            Putting , the given values -:

                [tex]125\times\frac{12}{120}[/tex]

         =   12.5

Therefore , the number of turns on secondary coil is 12.5 .

which one is odd copper,plastic,rubber​

Answers

Answer:

It's plastic.

trust me it's plastic, i've rad it somewhere.

All of them have something that's not like the others.

-- Rubber is the only one on the list that has two repeated letters.

-- Plastic is the only one on the list thagt has no repeated letters.

-- Plastic is the only one on the list that has no 'r' in its name.

-- Copper is the only one on the list that is an element, not a compound.

-- Copper is the only good electrical conductor on the list.

-- Plastic is the only one on the list with more than six letters in its name.

-- Rubber is the only one on the list with no 'p' in its name.

-- Plastic is the only one on the list that doesn't end in "-er".

Scientific work is currently under way to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that driv- ers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long expo- sure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 3 1023 T, oscillating sinusoidally at 60.0 Hz. If the diameter of a red blood cell is 8.00 mm, determine the maximum emf that can be generated around the perimeter of a cell in this field.

Answers

Answer:

fem = 7.58 10⁻⁵ V

Explanation:

For this exercise we use Faraday's law

          fem =   [tex]- \frac{d \Phi _B}{dt}[/tex]

the magnetic flux is

          Ф_B = B. A = B A cos θ

Tje bold are vectros.  Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1

The red blood cell area is

          A =π r²

       

indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m

the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T,  therefore we can write it

          B = B₀ sin (wt) = B₀ sin( 2π f t)

           

we substitute

           fem = - A dB / dt

           fem = - A B₀  [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]

           fem = - π r² Bo (2πf  cos 2πft)

the maximum electromotive force occurs when the function is ±1

       

           fem = 2 π² r² B₀ f

let's calculate

           fem = 2π²  (8.00 10⁻³)²  1.00 10⁻³ 60

           fem = 7.58 10⁻⁵ V

A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
kg, collides with a smaller box with a mass of 2 kg that is stationary. The boxes stick
together. What is the velocity of the two combined masses after collision?
8 m/s
O m/s
12 m/s
4 m/s
us 12:18

Answers

Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48

Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u


Mb=Ma
48=6u
u=8m/s

A train travels at a speed of 70km/h and travels a distance of 630 km. How long did it take the train to complete its journey?

Answers

Answer:

9 hr

Explanation:

Speed = distance/time

Let the time taken by the train be t.

=> 70 = 630/t

=> t = 630/70

=> t = 9

it take 9 hr to complete its journey

Answer:

9 hours

Explanation:

Using the formula to solve motion problems: d=r*t (Distance = Rate * Time)

Given:

Distance = 630 km

Rate (Speed) = 70 km/h

Time = Distance / Time

Time = 630/70

               = 9

uniform solid sphere has a mass of 1.765 kg and a radius of 0.854 m.a. Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s.b. What magnitude force applied tangentially at the equator would provide the needed torque

Answers

Answer:

a) the torque required is  10.53 N-m

b) The magnitude force applied tangentially is 12.33 N

Explanation:

Given the data in the question;

mass m = 1.765 kg

radius r = 0.854 m

first we calculate the moment of inertia;

[tex]I[/tex] = [tex]\frac{2}{5}[/tex]mr²

we substitute

[tex]I[/tex] = [tex]\frac{2}{5}[/tex] × 1.765 × (0.854)²

[tex]I[/tex] = 0.514897 kg.m²

a)

Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s

ω[tex]_{initial[/tex] = 0

ω[tex]_{final[/tex] = 317 rad/s

t = 15.5 s

we know that; ω[tex]_{final[/tex] = ω[tex]_{initial[/tex] + ∝t

so we substitute

317 = 0 + ∝(15.5)

∝ = 317 / 15.5

∝ = 20.4514 rad/s²

so

ζ = [tex]I[/tex] × ∝

we substitute

ζ = 0.514897 × 20.4514

ζ = 10.53 N-m

Therefore, the torque required is  10.53 N-m

b)

What magnitude force applied tangentially at the equator would provide the needed torque.

ζ = F × r

we substitute

10.53  = F × 0.854

F = 10.53 / 0.854

F = 12.33 N

Therefore, magnitude force applied tangentially is 12.33 N

A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
the cat's displacement during this motion?
A. 6m
B. 2.5 m
C. 8.5 m
D. 4.5 m
20
А
B
D​

Answers

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force on the sled by pulling on the rope. Suppose the coefficient of kinetic friction between the loaded sled and snow is 0.200.
(a) The Eskimo pulls the sled 5.90 m, exerting a force of 1.10 102 N at an angle of θ = 0°. Find the work done on the sled by friction, and the net work. Wfric = Correct: Your answer is correct. . J Wnet = Correct: Your answer is correct. . J
(b) Repeat the calculation if the applied force is exerted at an angle of θ = 30.0° with the horizontal. Wfric = J Wnet = J

Answers

Answer:

(a)

W_friction = 98.1 J

W_net = 550.9 J

(b)

W_friction = 98.1 J

W_net = 463.95 J

Explanation:

(a)

First, we will calculate the work done by friction:

[tex]W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\[/tex]

W_friction = 98.1 J

Now, the work done by Eskimo will be:

[tex]W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 0^o\\[/tex]

W_Eskimo = 649 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 649 J - 98.1 J

W_net = 550.9 J

(b)

First, we will calculate the work done by friction:

[tex]W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\[/tex]

W_friction = 98.1 J

Now, the work done by Eskimo will be:

[tex]W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 30^o\\[/tex]

W_Eskimo = 562.05 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 562.05 J - 98.1 J

W_net = 463.95 J

Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Erase button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one with 12 V, one with 15 V, and one with 6 V). Don't worry about getting these exact values. You can be off by a few tenths of a volt. Which statement best describes the distribution of the equipotential lines?
1. The equipotential lines are closer together in regions where the electric field is weaker.
2. The equipotential lines are closer together in regions where the electric field is stronger.
3. The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.

Answers

Answer:

B or 2

Explanation:

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