The electric potential, which is a scalar field, may be represented graphically by equipotential curves. The lines of electric field may be obtained from those equipotentials.

Four rectangles have been placed on the graph below. Assess the region of the graph where the magnitude of the electric field is greatest. Drag the label Emax to the rectangle in that region. Assess the region of the graph where the magnitude of the electric field is smallest. Drag the label Emin to the rectangle in that region.

The Electric Potential, Which Is A Scalar Field, May Be Represented Graphically By Equipotential Curves.

Answers

Answer 1

The region with high electric potential  (80 V and 70 V) will have high electric field while the region with low electric potential (20 V) will have low electric field.

What is Electric field?

Electric field is the field that surrounds electrically charged particles and exerts force on all other charged particles in the field.

Relationship between electric potential and electric field

E = V/d

where;

V is electric potential (V)E is electric field (V/m)d is the distance (m)

Since electric field is directly proportional to electric potential, the region with high electric potential  (80 V and 70 V) will have high electric field while the region with low electric potential (20 V) will have low electric field.

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Related Questions

A student in the Biomechanics class has decided that she would like to make her arms
stronger. She has a mass of 63 kg, She chooses to complete some elbow flexion exercises
using a kettlebell. For this problem, consider the hand and forearm to be a single segment.
The distance from her elbow to her wrist is 22.86 cm.
The force from the kettlebell is applied to her hand, which is 30.48 cm from her elbow joint.
She knows that the moment arm of the elbow extensor muscles about the elbow axis is

Answers

Answer:

what is heat and transfer

Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

Answers

Answer:

Gravitational force is the force that attracts objects towards each other. Two factors that affect the gravitational force between objects are the mass of the two objects and the distance between

Explanation:

Gravity is what pulls us towards the Earth if we were to jump into the air, so it is the force that pulls things towards other things. The bigger the objects are the more gravity they have, so a planet has more gravity than say, an apple. Distance between objects also makes their gravity change, so the Earth's pull on the moon is different than the Earth's pull on the sun.

Hopefully this helps- let me know if you have any questions!

Question :-

Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

Answer :-

Gravitational Force :-

It is the attractive force between the two body having some mass.Example of gravitational force :- attraction between the human and earth . earth pulls the body towards it with the help of this This force helps the celestial bodies to revolve in there orbits

Factors effecting Gravitational force:-

Mass :- gravitational force directly proportional on the masses of the body between which it is acting [tex]gravitational \: force ∝m{ \tiny1}m{ \tiny2}[/tex]Distance between the masses :- Gravitational force is inversely proportional to the square of the distance between the masses [tex]gravitational \: force∝ \frac{1}{ {r}^{2} } [/tex]

Formula of the Gravitational force :-

[tex]F = \frac{G m{ \tiny1}m{ \tiny2} }{ {r}^{2} } [/tex]

where

F is gravitational force G is gravitational constant ( universal constant ) m 1 is mass of first body m2 is mass of second body r is the distance between the two bodies

What is environmental?

Answers

Answer:

Environmental means relating to or caused by the surroundings in which someone lives or something exists. It protects against environmental hazards such as wind and sun. The form the human family takes is a response to environmental pressures.

Explanation:

I HOPE IT HELPS!!

THANX!!!

STAY SAFE!!!!

Find the temperature

Answers

Answer:

-------1

Explanation:

beacuse that is what i know

What’s the weight of a box w/ a mass of 150kg on earth?

Answers

[tex]\text{Weight ,} W = mg = 150 \times 9.8 =1470N[/tex]

Explanation:

f=ma

where force is weight. we know that a is the acceleration of gravity which is -9.8 m/s^2

so f is

-1471.5 N

Mention the objective of the Experiment?

Answers

Answer:

I don't understand your question ❓,the object.....of what experiment

The objective of this experiment is to learn whether any Brainly contributor can give a reasonable or helpful answer to a question that contains totally zero information.

7. A 2.0 kg block, starting from rest, is pushed by a
constant force along a frictionless track. The
position of the block as a function of time is
recorded in the data above. The final momentum
of the block is
(A) 0.8 kgm/s
(B) 1.2 kgm/s
(C) 1.6 kgm/s
(D) 3.2 kgm/s

Answers

Answer:

(A) 0.8 kgm/s

Explanation:

because of the even ground it would only slow down

Terry is walking down the street at 3 m/s. If he
has a mass of 70 kg, what is his momentum?

Answers

[tex]\text{Given that, mass m = 70 kg and velocity v = 3 m/s}\\\\\text{Momentum,}~ p = mv = 70 \times 3 = 210~ kg ~ms^{-1}[/tex]

As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?

Answers

As a conservation biologist for the Chesapeake Bay, you and your

colleagues have been conducting a research study that tracks the decrease

in the bald eagle population over the past few years.

What evidence can you find for the decrease in the bald eagle population?

Suppose a grower sprays (2.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water when it freezes?

Answers

There is no temperature change which drives heat flow, thus no heat will be released by the water.

Heat released by the water when it freezes

The heat released by the water when it freezes is calculated as follows;

Q = mcΔФ

where;

m is mass of waterc is specific heat capacity of waterΔФ is change in temperature = Фf - Фi

Initial temperature of water, Фi = 0 °C

when water freezes, the final temperature, Фf = 0 °C

Q = 22 x 4200 x (0 - 0)

Q = 0

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

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what is one type of compact star with a mass similar to the sun but a diameter similar to earth?

Answers

Explanation:

neutron star, any of a class of extremely dense, compact stars thought to be composed primarily of neutrons. Neutron stars are typically about 20 km (12 miles) in diameter. Their masses range between 1.18 and 1.97 times that of the Sun, but most are 1.35 times that of the Sun.

A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per meter of length. By constructing a suitable Gaussian surface around the wire, Find the magnitude and direction of the electric field at points

a)5.5m

b)2.5m

perpendicular from the center of the wire.​

Answers

(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.

(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.

Electric field at a point on the Gaussian surface

The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;

E = λ/2πε₀r

where;

λ is linear charge densityε₀ is permitivity of free spacer is the position of the chargeAt a distance of 5.5 m

[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C[/tex]

At a distance of 2.5 m

[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C[/tex]

Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.

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Why would it be impractical to wire a home with a circuit in which all loads were connected in series.

Answers

For residences, series circuits are impracticable. Parallel circuits are practical because each appliance is controlled by its own switch, which prevents the other appliances from being turned off. A) The appliances are all turned on. The conductor wire is carrying a considerable quantity of current (arrow).

list out the use of simple machine​

Answers

Explanation:

simple machine can multiplayer of speed and force

A 4500 kg Aston Martin traveling at 102 m/s has to stop short because some ducklings
hazard onto the road. The Aston Martin was able to stop in 1.77 seconds. How much
force was placed on the car?

Answers

Answer:

-259322.03N

Explanation:

[tex]F=m*(\frac{v}{t})\\ F=4500kg*(\frac{0-102m/s}{1.77s} )\\F=-259322.033898\\\\[/tex]

a
Which of these is a chemical
change?
A. water boiling
B. salt disolving
C. paper burning

Answers

Answer:

Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.

Explanation:

Two identical charges are located 1 m apart and feel a 1 N repulsive electric force. What is the charge of each particle.

Answers

The charge on each particles which are 1 m apart and feeling a repulsive force of 1 N is 1.05×10¯⁵ C

Assumption

Let the charge on each particles be q

How to determine the charge Final force (F) = 1 NDistance apart (r) = 1 mElectrical constant (K) = 9×10⁹ Nm²/C²Charge on 1st particle (q₁) = q =? Charge on 2nd particle (q₂) = q =?

The charge on each particle can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = Kq² / r²

1 = (9×10⁹ × q²) / 1²

1 = 9×10⁹ × q²

Divide both side by 9×10⁹

q² = 1 / 9×10⁹

Take the square root of both side

q =  √(1 / 9×10⁹)

q = 1.05×10¯⁵ C

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An electron has a mass of 9.1x10^-31 kg and a charge
of -1.6x10^-19 C. Suppose you could isolate one electron in
a perfect vacuum and then create an electric field to pull
and edi 107. upward on the electron. How strong would the field have to
be to counteract the electron's weight? (In other words,
di Delfine bu how strong would the field have to be to put the electron in
a state of force equilibrium?)

Answers

is one of these the question

Why was the government in giving support to addressing and researching the HIV epidemic ?

Answers

Do you mind adding the picture?? It’ helps me answer the question better :)

The force of gravity on an object gives the object its _______.
A. mass
B. kilogram
C. acceleration of gravity
D. weight

Answers

Answer:

Weight

Explanation:

The force of gravity on a an object gives the object its weight.

a=5i+4j-6k ,b=-2i+2j+3k ,c=4i+3j+2k. find the vector perpendicular to a and c​

Answers

Answer:

Explanation:

You can use the cross product. Let the vector that perpendicular to a and c is [tex]\vec{d}[/tex], so:

[tex]\vec{d}=\vec{a}\times\vec{c}=\left|\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\5&4&-6\\4&3&2\end{array}\right] \right|=(8+18)\hat{i}-\hat{j}(10+24)+\hat{k}(15-16)=26\hat{i}-34\hat{j}-\hat{k}[/tex]

To check that c is perpendicular with a and b, do the dot product between c and a and also c and b and if the result is zero, you're true.

[tex]\vec{d}.\vec{a}=(26*5)-(34*4)+(6)=0[/tex]  (c perpendicular to a)

[tex]\vec{d}.\vec{c}=(4*26)-(34*3)-(2*1)=0[/tex] (d perpendicular to c)

A rope is wrapped around a pulley many times. The pulley can be modeled as a solid disk of radius R and mass M, and a mass mA hangs vertically from the pulley. The mass is released from rest. show answer Incorrect Answer 25% Part (a) What is the magnitude of the tangential acceleration of the hanging mass?

Answers

The magnitude of the tangential acceleration of the hanging mass is 2mg/MR

Tangential acceleration of the hanging mass

The tangential acceleration of the hanging mass around the pulley is determined from the principle of conservation of angular momentum as shown below;

τ = Iα

Where;

I is the moment of inertiaα is the angular velocity

[tex]\alpha = \frac{\tau}{I} \\\\\alpha = \frac{mgR}{3/2MR^2} \\\\\alpha = \frac{2mgR}{3MR^2} \\\\\alpha = \frac{2mg}{3MR}[/tex]

Where;

m is the hanging massM is the mass of solid disk

The tangential acceleration is calculated as follows;

[tex]a = \alpha R\\\\a = \frac{2mg}{3MR} \times R\\\\a = \frac{2mg}{3M}[/tex]

Thus, the magnitude of the tangential acceleration of the hanging mass is 2mg/MR

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2. A tennis ball machine launches balls horizontally with an initial speed of 5.3 m/s, from a height of 1.2 m.
a) What will the time of flight be for a tennis ball launched by the ball machine? (3)
b) What will the range of the tennis ball be? (2)
c) What will be the final velocity of the ball with which it reaches the ground? (3)

Answers

(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.

(b) The range of the tennis ball be is 1.01 m.

(c)  The final velocity of the ball with which it reaches the ground is 7.16 m/s.

Time of flight of tennis ball

The time of flight of the tennis ball is calculated as follows;

h = vt + ¹/₂gt²

1.2 = 5.3t + 0.5(9.8)t²

1.2 = 5.3t + 4.9t²

4.9t² + 5.3t - 1.2 = 0

a = 4.9, b = 5.3, c = 1.2

solve using quadratic formula

t = 0.19 s

Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.

Range of the tennis ball

The range of the tennis ball is calculated as follows;

R = vt

R = 5.3 x 0.19

R = 1.01 m

Final velocity of the ball

The final velocity of the ball with which it reaches the ground is calculated as follows;

vf = vo + gt

vf = 5.3 + 9.8(0.19)

vf = 7.16 m/s

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Determine
i. the total capacitance for the circuit
ii. the total charge stored in the circuit
iii. the charge stored in C9 (3μF)​

Answers

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is  6 x 10⁻⁶ C.

Total capacitance of the circuit

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

C1 and C2 are in series

[tex]\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F[/tex]

C1 and C2 are parallel to C3

[tex]C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F[/tex]

C(123) is series to C5 and C6

[tex]\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F[/tex]

C7 and C8 are in series

[tex]\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F[/tex]

Total capaciatnce of the circuit

Ct + C(78) = 2 μF + 3 μF = 5 μF

Total charge stored in the circuit

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

Charge stored in 3μF capacitor

Q =  (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

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The ________ of an object is the same on the earth as it is on the moon.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight

Answers

Answer:

mass

Explanation:

The mass of an object is the same on the earth as it is on the moon.

Mass always remains constant, it never changes with respect to place.

17) Which object would likely have the greatest velocity
a. a bouncy ball
b. a bowling call
a go-kart
d. a school bus

Answers

Answer: b or d

Explanation: b or d

Answer:

a

Explanation:

F= ma

interestingly

when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases

(man, PHYSICS IS JUST THE BESY)

A: a

object with the smallest mass has largest acceleration

An unbalanced force acting on an object results in ______.
A. inertia
B. friction
C. acceleration
D. faster
E. slower

Answers

Answer:

friction

Explanation:

A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
the temperature of the water rose by 4.20 ºC. If the heat capacity of the bomb plus water was 10.4 kJ/ºC,
calculate the molar heat of combustion of methanol.

Answers

Mass of methanol = 1.922g; Change in temperature = 5.14° C; Heat capacity of the bomb calorimeter + water = 8.69kJ/°C. Number of moles.

A cable with 19.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after being lifted 2.00 m ? Solve this problem using work and energy

Answers

The final speed of the block, after being lifted 2.00 m is 3.39 m/s

What is speed?

Speed can be defined as the rate of change in the distance of a body.

To calculate the speed of the block after being lifted 2.00 m,  first, we need to calculate the acceleration of the block using the formula below

Formula:

T-mg = ma......... Equation 1

Where:

T = Tension in the cablem = mass of the cablea = accelerationg = acceleration due to gravity

Restructuring the formula above,

a = (T-mg)/m............... Equation 2

From the question,

Given:

T = 19 Nm = 1.5 kgg = 9.8 m/s²

Substitute these values into equation 2

a = [(19)-(1.5×9.8)]/1.5a = 4.3/1.5a = 2.87 m/s²

Finally, to calculate the speed of the block, we use the formula below.

v² = u²+2as.......... Equation 3

Where:

v = Final speedu = initial speeda = accelerations = distance

From the question,

Given:

u = 0 m/sa = 2.87 m/s²s = 2.00 m

Substitute these values into equation 3

v² = 0²+(2×2×2.87)v² = 11.48v = √11.48v = 3.39 m/s

Hence, The final speed of the block, after being lifted 2.00 m is 3.39 m/s.

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A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed 4.20 m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 8.40 block recoils with a speed of 0.400 m/s. In the figure; the blocks are in contact for 0.200 s.

Answers

For A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed of 4.20 m/s  is mathematically given as

F = 193.2N

What is the magnitude of the average force on the 8.40-kg block, while the two blocks are in contact, is closest to?

Generally, the equation for the  magnitude of the average force mathematically given as

F = m(v1+v2)/t

F = 8.40(4.2+O.4)/t

F = 193.2N

In conclusion magnitude of the average force is

F = 193.2N

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