Which two factors does the power of a machine depend on?
OA. work and distance
OB. force and distance
OC. work and time
OD. time and distance

Answer:

Correct option is (A).

Explanation:

Given data:

Volume, [tex]V = 3 \ cm^{3}[/tex]

Mass, [tex]m = 8.1 \ g[/tex]

The density of the aluminium can be given as,

[tex]d = \frac{m}{V}[/tex]

[tex]d = \frac{8.1 \ g}{3.0 \ cm^{3}}[/tex]

[tex]d = 2.7 \ g/cm^{3}.[/tex]

Answer:The answer is tennis ball

Explanation:

i know this cause I'm good at math and why wouldn't you trust a random guy off the internet.

The refractive index of the liquid is 1.52. The substance is a crown glass.

Explanation:

Given:

Velocity of light through the substance, v = 1.97 X 10⁸ m/s

Refractive index of the liquid, μ = ?

We know,

μ = [tex]\frac{c}{v}[/tex]

where, c = velocity of light in air

            v = velocity of light in the substance

speed of light in air is 3 X 10⁸m/s

[tex]= \frac{3 X 10^8}{1.97 X 10^8} \\\\= 1.52\\[/tex]

Therefore, refractive index of the liquid is 1.52. The substance is a crown glass.

n = 1.52

Index of the liquid = 1.52

The type of the liquid is Crown glass.

Explanation:

In finding the index of refraction, we use this formula

n = c/v

where n is the index of refraction

c is velocity of light in vacuum which is constant and is equivalent to [tex]3 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

v is the velocity of light in a particular medium

Here we use [tex]1.97 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

Using the formula,

n = c/v

n = [tex]3 \times 10^{8} / 1.97 \times 10^{8}[/tex]

n = 1.52284264

n = 1.52

Thus, the index of the liquid is 1.52

Answer:

Hello, the tripping of a 230-kilovolt transmission line.

Explanation:

the tripping of a 230-kilovolt transmission line near Ontario, Canada, at 5:16 p.m., which caused several other heavily loaded lines also to fail. Hopefully this helps you find what your looking for!.

The case considered is of a kerosene lamp. Kerosene acts as a fuel to generate light. It undergoes combustion where it reacts with oxygen in the air to release light. The wick draws the kerosene and burns it. To increase the amount of time the lantern provides, we can decrease the length of the wick to avoid more consumption of the kerosene. Before lighting the wick, we first need to check the quantity of the kerosene present and also ensure to change the wick at least once a year.

Final answer:

To increase the burn-time of a lantern, one could use a larger kerosene supply, a more efficient wick, higher-grade kerosene, and ensure adequate oxygen supply. Similar to how argon extends the life of incandescent bulbs, optimizing lantern conditions prolongs its light duration.

Explanation:

When the wick of a lantern is lit, the kerosene undergoes a chemical reaction with oxygen, producing light through the release of energy. To increase the amount of time the lantern provides light, one can replace the consumed kerosene with a larger supply or use a wick that delivers the kerosene more slowly to the flame. Another method involves using a higher-grade kerosene that burns more efficiently, thus slowing down the rate at which it is used up.

Additionally, ensuring that the lantern is kept in an environment with adequate oxygen supply will prevent the flame from going out prematurely. Reflecting on the provided information, we can also look at how inert gases in light bulbs function to increase efficiency and longevity; similarly, optimizing the conditions for the lantern's operation can extend its burn time.

Just as argon helps to extend the life of incandescent lamp bulbs, taking measures to prevent rapid consumption of kerosene will extend the burn-time of a lantern. In light bulbs, argon prevents tungsten atoms from blackening the bulb and concentrates them back onto the filament; a parallel in lanterns would be maintaining the purity and functionality of the kerosene and wick.

Hay

Explanation:

Harvesting and storing forage at the proper moisture concentration is important for producing a high-quality product. Ideally, hay forage ought to have a moisture concentration between 15% to 18% during baling. Hay baled at higher moisture levels is liable to heat damage, dry-matter loss, mold spoilage, and hay fires. Protein and total digestible nutrient losses are increased for hay baled at lower moisture levels. Baling at the proper moisture level is critical to making quality hay, especially for larger bales (round and square) since moisture and heat dissipate gradually in contrast to smaller bales.

Hay is the type of forage that is cut and dried to contain low moisture, which is produced to store and preserve the feed for animals.

The type of forage that is cut and dried until it contains low moisture is known as hay. Haymaking is the process that transforms fresh, green forage into hay, which involves cutting, drying and then storing the plants. The purpose of creating hay is to preserve and store the forage at a nutritional state that is suitable for feeding animals, especially during times when fresh forage is not available, such as winter. The drying process is crucial as it prevents the growth of mold and spoilage organisms, ensuring the hay remains safe and suitable to feed livestock.

Answer: The force of a magnetic field on a particle with spin causes the particle to rotate it's spin to align with the magnetic field. ... Taking these two ideas together, then, the Earth's magnetic field will cause the magnet to align north to south. That's why people named them the north and south poles.

Explanation: Hope i could help please mark as brainliest!

Answer: 1 dioptre of power of a lens is defined as the unit of measuring the power of optical lens Or curved mirror which is equal to the reciprocal of the focal length. The focal length is measured in the meter. 1 dioptre is equal to 1/m where m is the focal length. Basically dioptre is the SI unit of optical power of the lens.

Answer:

B.

It will be greater than 10 J.

Explanation:

The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):

E = PE + KE

According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.

The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.

The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.

Here at the beginning, when it is at the top of the roof, the baseball has:

PE = 120 J

KE = 10 J

So the total energy is

E = 120 + 10 = 130 J

As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).

Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.

Answer:

Explanation:

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

[tex]F=mg[/tex]

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

[tex]g=\frac{GM}{(R+h)^2}[/tex]

where:

G is the gravitational constant

[tex]M=5.98\cdot 10^{24} kg[/tex] is the Earth's mass

[tex]R=6.37\cdot 10^6 m[/tex] is the Earth's radius

Here,

[tex]h=6.38\cdot 10^6 m[/tex]

So the acceleration due to gravity is

[tex]g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2[/tex]

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

[tex]F=mg=(70)(2.45)=171.5 N[/tex]

The gravitational force on a 70 kg person at the Earth's surface can be calculated using Newton's law of universal gravitation by inputting the mass of the Earth, the person's mass, and the radius of the Earth into the gravitational force formula.

To calculate the gravitational force on a person 70 kg in mass and 6.38 x 106 meters above the Earth's surface, we use Newton's universal law of gravitation. The formula for gravitational force (F) is: F = GmM/r2, where G is the universal gravitational constant, m is the mass of the person, M is the mass of the Earth, and r is the distance from the person to the center of the Earth.

The value of universal gravitational constant (G) is 6.674 x 10-11 N·m2/kg2. The mass of the Earth (M) is 5.97 x 1024 kg. Given these values and the person's mass (m), the gravitational pull on them can be calculated.

The distance to the center of Earth (r) would be the sum of Earth's radius and the altitude above Earth's surface, which in this case is simply Earth's radius, as the person is on the surface. Substituting the values into the formula gives F = (6.674 x 10-11 N·m2/kg2 x 70 kg x 5.97 x 1024 kg) / (6.38 x 106 m)2 to find the gravitational force.

a) 1 A° corresponds to 1.22 B°

b) [tex]105.4^{\circ}B[/tex]

Explanation:

a)

The image of the problem is missing: find it in attachment.

Here we have two temperature scales:

- In scale A, the freezing point is at [tex]-30.0^{\circ}A[/tex], while the boiling point is at a temperature of [tex]+60.0^{\circ}A[/tex]

- In scale B, the freezing point is at [tex]+20.0^{\circ}B[/tex], while the boiling point is at a temperature of [tex]+130.0^{\circ}B[/tex]

The freezing point and the boiling point must be at the same absolute temperature on the two scales: this means that the interval

(boiling point - freezing point)

must be the same in the two scales.

For scale A, this interval is equal to:

[tex]\Delta T_A = +60.0-(-30.0)=90.0^{\circ}A[/tex]

For scale B, this interval is equal to:

[tex]\Delta T_B=+130-(+20)=110^{\circ}B[/tex]

Therefore, 1 degree on the A scale corresponds to:

[tex]\frac{x^{\circ}B}{1^{\circ}A}=\frac{110^{\circ}B}{90.0^{\circ}A}\\x^{\circ}B=\frac{110}{90}\cdot 1=1.22[/tex]

So, 1 A° corresponds to 1.22 B°.

b)

The temperature of the substance on the A scale is

[tex]T_A=+40.0^{\circ}A[/tex]

This means that the difference of its temperature relative to the freezing point on the A scale is

[tex]\Delta T_A=+40.0-(-30.0)=70.0^{\circ}A[/tex]

In part A, we stated that 1 degree on the A scale corresponds to 1.22 degrees on the B scale, so this temperature interval on the B scale is

[tex]\Delta T_B=1.22\Delta T_A=1.22(70)=85.4^{\circ}B[/tex]

And therefore, the temperature of the substance on the B-scale is equal to the freezing point (on the B scale) plus this interval:

[tex]T_B=+20.0+85.4=105.4^{\circ}B[/tex]

To convert temperature readings from scale A to scale B, find the ratio of the differences between the boiling and ice points on each scale, then use this ratio to multiply the temperature on scale A.

The subject matter of the question is temperature conversion between two different scales - scale A and scale B. To answer the questions:

For example, if the boiling point of water is 100A and 80B, and the ice point is 0A and 20B, the calculation would be (80B-20B)/(100A-0A) = 0.6 B/A. Therefore, if a temperature reads +40.0A, it would read 0.6*40=24B on the B scale.

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Answer:

a is correct answer

Explanation:

proton protons

Final answer:

The atomic number refers to the number of protons in an atom's nucleus and is crucial in identifying an element. The correct answer is A) proton, protons.

Explanation:

The atomic number, or proton number, is described as the number of protons in the nucleus of a chemical element. Thus, the correct answer to the question is:

A) proton, protons

The atomic number (Z) is a unique identifier for each element, determining its identity based on the number of protons in its nucleus. For example, carbon has an atomic number of 6, meaning all carbon atoms have six protons in their nucleus. A neutral atom has an equal number of protons and electrons. The sum of the number of protons and neutrons is referred to as the mass number (A), and isotopes of an element will have the same atomic number but can differ in the number of neutrons.

Answer:8NS

Explanation:

Impulse=change in momentum

Impulse=mv-mu

Impulse=4x12-4x10

Impulse=48-40

Impulse=8NS

Answer:

Explanation:

As with all atoms known, electrons moves in discrete energy levels round a well defined orbit.

Protons are restricted to the nucleus of an atom. They occupy this small and massive position along with the neutrons.

Metals, have a large dense cloud of electrons.

Therefore, it is safe to conclude that in metal conductors, electrons move but protons do not just like in every other atom.

Protons do not move in a metal conductor because they are stable in the nucleus.

A conductor is a type of material that allows the flow of charge (electric current) in one or more directions.

In metallic conductors, the movable charged particles are electrons, though in other cases they can be ions.

Every atom of a conductor consists of three particles protons, neutrons and electrons.

Protons are attracted to both other protons as well as neutrons. So inside a nucleus, there exists a battle between electrical repulsion and nuclear attraction. The proton by itself is stable but the strong force is not quite strong enough to bind two protons.

Hence we can conclude that in a metal conductor, electron moves but protons do not move.

To know more about the conductors, follow the link given below.

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Answer:

1)

When the person throws the ball away, the person rolls backward. This is due to the law of conservation of momentum: in fact, the total momentum of the person+ball system must be conserved.

At the beginning,

[tex]p_i=0[/tex]

after throwing the ball, the total momentum is the sum of the momentum of the person and of the ball:

[tex]p_f=p_p + p_b[/tex]

Since momentum is conserved,

[tex]p_i = p_f\\0=p_p+p_b[/tex]

So

[tex]p_p = -p_b[/tex]

Therefore, the person has equal momentum (in magnitude) but opposite direction to the ball, so the person rolls backward.

However, if the person hold to the ball, then they will have same momentum (moving in the same direction). In order to conserve the total momentum (which was zero at the beginning), the only possible solution is that

[tex]p_p=p_b=0[/tex]

which means that both the person and the ball will remain at rest. This is because there are no external forces acting on the system, so the system cannot move.

2)

The change in momentum of an object is given by

[tex]\Delta p=m(v-u)[/tex]

where

m is the mass of the object

v is its final velocity

u is the initial velocity

For the clay ball in this problem, we have:

m = 50 g = 0.050 kg

v = 0 m/s (it sticks on the wall)

u = 1 m/s

So its change in momentum is

[tex]\Delta p_c=(0.050)(0-1)=-0.050 kg m/s[/tex]

For the superball, we have:

m = 50 g = 0.050 kg

v = -0.8 m/s (it bounces back)

u = 1 m/s

So its change in momentum is

[tex]\Delta p_s = (0.050)(-0.8-1)=-0.09 kg m/s[/tex]

So, the superball has a greater change in momentum (in magnitude).

3a)

According to Newton's third law of motion:

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A".

Here, we have a Hummer and a Beetle colliding head-on: we can identify them as object A and object B. Therefore, according to Newton's third law:

- The action force is the force of impact exerted by the Hummer on the Beetle

- The reaction force is the force of impact exerted by the Beetle on the Hummer

And according to the Law, the two forces are equal in magnitude: so, the two vehicles experience the same force of impact.

3b)

The change in momentum of each vehicle during the collision can be written as

[tex]\Delta p = F\Delta t[/tex] (1)

where

[tex]\Delta p[/tex] is the change in momentum

[tex]F[/tex] is the force experienced by the vehicle

[tex]\Delta t[/tex] is the duration of the collision

in part 3a), we said that the two vehicles experience the same force in the collision.

Moreover, the duration of the collision, [tex]\Delta t[/tex], is the same for the two vehicles.

As a result, according to formula (1), the two vehicles have same change in momentum (however, the directions would be opposite, since they experience force in opposite directions).

3c)

According to Newton's second law of motion, the acceleration of an object is given by:

[tex]a=\frac{F}{m}[/tex]

where

F is the force experienced by the object

m is its mass

a is its acceleration

In part 3a), we stated that the force experienced by the Beetle and the Hummer is the same. However, the mass of the Beetle is smaller than the mass of the Hummer: from the equation we see that the acceleration is inversely proportional to the mass, therefore the Beetle will experience a greater acceleration.

4a)

The force experienced by the dashboard on the car is given by:

[tex]F=\frac{\Delta p}{\Delta t}[/tex]

Where

[tex]\Delta p[/tex] is the change in momentum

[tex]\Delta t[/tex] is the duration of the collision

In a padded dashboard, the duration of the collision [tex]\Delta t[/tex] is larger than the duration of the collision for a hard dashboard. According to the equation above, the force experienced by the dashboard (and so, the car) is inversely proportional to the duration of the collision: therefore, since the padded dashboard has a larger [tex]\Delta t[/tex], it will experience a smaller force than the hard dashboard.

4b)

The force experienced by the climber if falling is given by

[tex]F=\frac{\Delta p}{\Delta t}[/tex]

Where

F is the force experienced by the climber

[tex]\Delta p[/tex] is his change in momentum

[tex]\Delta t[/tex] is the duration of fall

Nylon is a very elastic material, so it is able to "soften" the fall by stretching a lot. As a result, the nylon increases the value of [tex]\Delta t[/tex] in the formula. Since the force experienced by the climber is inversely proportional to [tex]\Delta t[/tex], the climber will feel less force thanks to the nylon.

4c)

This technique is used to exploit the "push" given by the second car of the train to the first car when the brakes are applied.

At first, the engine is started, and the first car starts accelerating, pulling the second car (and the following cars). Then, the brakes are applied on the first car: however, the second car keeps moving by inertia, so then it gives a push forward on the first car. Then, this action is repeated several times, so that this push exerted by the second car is exploited several times.

Answer:

It is B

it is called a jumper cable because it jumps the power from one car to the other

The name for the thick, insulated cables used for jump starting a car is "jumper cables".

Jumper cables, which are sometimes referred to as booster cables and jump leads, are both a pair of insulated wires with just enough capacity and clamp at each end to communicate the disabled vehicle or equipment with such an auxiliary source, including another vehicle or piece of equipment with about the same system voltage or even to a different battery.

A non-conductive material that is resistive to an electric current makes up an insulated wire and cable. It encloses as well as safeguards the cable and perhaps wire inside.

The name for the thick, insulated cables used for jump starting a car is "jumper cables".

Therefore, the correct answer will be option (B).

To know more about Jumper cables

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Celsius . . . negative 273.15 degrees

Kelvin . . . . zero of them

Final answer:

The lowest temperature on the Celsius scale is -273.15 °C and on the Kelvin scale is 0 K, both representing absolute zero. The kelvin is an absolute temperature unit used in science, with the freezing and boiling points of water at 273.15 K and 373.15 K respectively.

Explanation:

The lowest possible temperature on the Celsius scale is -273.15 °C, which is known as absolute zero. On the Kelvin scale, the lowest possible temperature is 0 K, also representing absolute zero. Absolute zero is the temperature at which all molecular motion ceases, making it the point where no thermal energy is produced by molecular motion. As the Kelvin scale is an absolute temperature scale used extensively in scientific work, it defines 0 K as the point of absolute zero. Unlike other temperature scales, it has no negative values, and the official temperature unit of this scale is the kelvin, simply abbreviated as K without a degree sign.

The Kelvin scale's importance stems from many physical quantities, such as the volume of an ideal gas, being directly related to absolute temperature. The Celsius and Kelvin scales are aligned such that the freezing and boiling points of water are 273.15 K and 373.15 K, respectively, meaning that the magnitude of temperature differences is the same in units of kelvins and degrees Celsius.

Answer:

I = 4.28 [amp]

Explanation:

To solve this type of problems we must have knowledge of the law of ohm, which tells us that the voltage is equal to the product of resistance by current.

Initial data:

v = 1.5 [volt]

R = 0.35 [ohms]

v = I * R

therefore:

I = 1.5 / 0.35

I = 4.28 [amp]

Final answer:

Using Ohm's Law, the current in the circuit with a voltage of 1.5 volts and a resistance of 0.35 ohms is calculated to be approximately 4.286 amperes.

Explanation:

To determine the current in the circuit, we can use Ohm's Law, which is stated as I = V/R, where I is the current in amperes, V is the voltage in volts, and R is the resistance in ohms. Given that the voltage (V) is 1.5 volts and the resistance (R) is 0.35 ohms, we can substitute these values into the equation to calculate the current.

Thus, the current (I) can be calculated as follows:

I = 1.5V / 0.35Ω = 4.286 amperes (approx).

This means that the current flowing through the circuit is approximately 4.286 amperes.

Answer: Point A

Explanation: Gravitational potential energy is the energy that an object has because of its height and is equal to the object’s mass multiplied by it’s height multiplied by the gravitational constant (PE = mgh). Gravitational potential energy is greatest at the highest point of a roller coaster and least at the lowest point.

Answer:

Point A

Explanation:

Answer:

40.6 N/m

Explanation:

From Hooke's law we relate the extension of a spring and the applied force by the formula F=kx where F is the applied force, k is spring constant and x is extension of spring.

Making k the subject of the formula then we have

[tex]k=\frac {F}{x}[/tex]

We also know that F=gm where g is acceleration due to gravity which we take as 9.81 m/s2 while m is the mass. Substituting F with gm into the formula we get that

[tex]k=\frac {gm}{x}[/tex]

Substituting m for 5.8 kg and x for 1.4 m we get the spring conatant as

[tex]k=\frac {9.81\times 5.8}{1.4}=40.641428571428\ N/m\approx 40.6\ N/m[/tex]

Therefore, the spring constant is approximately 40.6 N/m

Answer:

The object will travel at a constant rate in along a straight line.

Explanation:

In the given situation, it is mentioned that there is no external force acting on the given object. Thus, it will retain its initial velocity along a straight path.

An object given an initial velocity and no other forces acting on it will move at a constant rate in a straight line.

The object will move at a constant rate in a straight line.

If an object is given an initial velocity and no other forces act on it, it will continue to move at a constant speed in a straight line. This concept is a fundamental principle in physics known as Newton's first law of motion or the law of inertia. According to this law, an object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force. In reality, there are usually forces like gravity and friction that can affect an object's movement, but, in theory (or in a vacuum), without those forces, an object's motion would remain unchanged.

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Answer:

2 × 10⁸J/kg

Explanation:

Specific latent heat of vaporization (steam)  is the quantity of heat required or needed to change a unit of mass of liquid to gaseous state at constant temperature (boiling point) and pressure.

It can be calculated mathematically using,

Q = mL

Where,  Q = quantity of heat

m= mass in kg

L = specific latent heat (J/kg)

Therefore,from the question

Q= 15000000J

mass (m)  = 75g = 75/1000 kg = 0.075kg

Q = mL

L = Q/m

L = 15000000  ÷ 0.075

L= 200000000J/kg

L = 2 × 10⁸J/kg

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sticky tape, increase, sticky thing cause some other things to stop

oil, decreas, oil is one of the slipprey things there is, so it would not help increase fricktion

water, again it will decrease

sand, it would decrease ,, making things difucoult to slide

wax, it would increase

work gloves would increase

hope this helps!

~ angilina

Final answer:

Common items like sticky tape, oil, water, sand, wax, work gloves, and grease can be used to either increase or decrease friction depending on the situation.

Explanation:

The common items can be used to either increase or decrease friction depending on the situation. Sticky tape can be used to increase friction when applied to surfaces to make them stick together. Oil, grease, and wax are commonly used as lubricants to decrease friction between surfaces, making them smoother and reducing resistance. Water and sand can be used to increase friction in certain situations, such as applying water to a surface to make it less slippery or using sandpaper to roughen a surface.

Answer:

.

Explanation:

Answer:

d. A voltage causes electric charges to move from higher to lower potential.

Explanation:

Answer:

Explanation:

Weight is the earth pull of an object to itself.

The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity.

The unit of weight is Newton and it is a vector quantity, it is directed downward vertically.

W= mg.

Where g varies but the gravity on earth is 9.81m/s² approximately.

Given that,

m=10kg, g=9.81

Then,

W=mg

W=10×9.81

W=98.1N

Answer:  

To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons. Example: Find the atomic mass of an isotope of carbon that has 7 neutrons. You can see from the periodic table that carbon has an atomic number of 6, which is its number of protons.

Explanation:

Answer:

A

Explanation:

kepler's research was based on the data and observations of tycho brahe.

Answer:

I would strongly recommend Exploring Quantum Physics through Hands-on Projects for physics practicals.

Explanation:

The best book for physics practicals, particularly for university-level students, is often considered to be 'Principles and Practice of Physics' by Eric Mazur due to its clear examples and structured sequence. OpenStax's 'University Physics I' is an excellent alternative with open-source availability. A balance of textbooks, resources, and instructor guidance can provide a robust physics learning experience.

When searching for the best book for physics practicals, the University Physics I by Eric Mazur, entitled Principles and Practice of Physics, comes highly recommended. This book is known for its clear explanations, terminology consistency with other resources, and its sequence of content presentation. It is packed with worked-out examples and offers self-quiz conceptual questions that are crucial for understanding practical aspects of physics.

Alternatively, students looking for online and open-source textbooks can turn to resources like University Physics I at OpenStax, which offers an abundance of exercises and a more traditional presentation style. For computational physics enthusiasts, texts like A Practical Introduction to Computational Physics and Scientific Computing could be explored.

Ultimately, these books along with different ways to digest physics concepts, such as discussions with instructors and accessing resources in the Physics library, can cater to the diverse needs of physics students. It's important to consider one's own learning style and curriculum requirements when choosing the right practical physics text.

Answer:

The correct statements  are:

Explanation:

The answer choices are:

The electric field is the electrostatic force per unit of charge,  

         [tex]\vec E=\dfrac{\vec F}{Q}[/tex]

around around a charge, where another charge would experience the electrostatic force.

The electric field lines are shown in a diagram with arrows ditributed radially away from a positive charge and radially toward a negative charge.

Since the arrows are away from Q, Q is a positive charge: statement D.

Since the size of the arrows decreases as you move away  from Q the stregth of the field is not uniform: statement A.

Since the charge Q is positive, a negative charge would be attracted toward it: statement E.

Answer:

Explanation:

Field lines are lines of forces around a bar magnet. The show the direction of force field in a magnet.

Usually, around a bar magnet, the field lines originates and spreads out from the north pole.

Then they converge and enter through the south pole.

Therefore, we can make our choice by inspecting the given diagram. If the lines enters through the pole, it is the south pole.

Where they originate or leave is the north pole.

In a magnet, the lines (forces) always enter through the south pole and leave through the north pole.

I hope that helps ya'll.

- sincerelynini