How has the magnet affected the block of material next to it?
A. The domains in the end of the block farthest from the magnet are aligned with its magnetic field.
B. The domains in the end of the block closest to the magnet are aligned with its magnetic field.
C. The domains in the center of the block are aligned with the magnetis magnetic field
D. The domains in the entire block are aligned with the magnets magnetic field​

How Has The Magnet Affected The Block Of Material Next To It? A. The Domains In The End Of The Block

Answers

Answer 1

Answer: (APE X) B

Explanation: I am taking it and I just got it right

Answer 2

The magnetic characteristics of a substance are determined by the magnetic moments of its atoms,  the domains which are closest to the magnet are aligned with its magnetic field, hence option b is correct.

What is the effect of a magnet?  

Long-range alignments of magnetic moments in ferromagnetic materials, known as domains, include magnetic moments that all point in the same direction.

This indicates that the atoms' individual magnetic moments are lined up with one another and pointing in the same general direction.

Therefore, a magnetic substance will get magnetized if it is placed close to a magnet. Magnetism is the name for this kind of electricity, hence option b is correct.      

Learn more about the magnet, here:

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Related Questions

A school bus's velocity decreases from 15 m/s to 5 m/s forwards over a period of 10 s. What is the bus's average acceleration? [Vf=v,+at]

Answers

Answer:-1 m/s/s

Explanation:

The acceleration of the bus when its velocity changes from 15 m/s to 5 m/s will be a=1 m/s^2

What is acceleration?

Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction.

Now from the first equation of motion we have,

[tex]a= \dfrac{v-u}{t}[/tex]

we have V=15 m/s, u=5 m/s , t=10 seconds

[tex]a=\dfrac{15-5}{10}=1\ \frac{m}{s^2}[/tex]

Hence the acceleration of the bus will be a=1 m\s^2

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a girl weighing 350N runs up hill upto 5m in 20sec. calculate her work done and power

Answers

P=350/300+20. P=350/320. P=35/32. P=1.09

Work done against gravity becomes potential energy of the object-Earth system. Its given by:

W = Ep

W = mgh  but weigh(w) = mg

W = wh

W = 350 N × 5 m

W = 1750 J

Power is the amount of energy transferred or converted per unit time. This is:

P = W/t

P = 1750 J / 20 s

P = 87.5 W

A
B
с
4 teaspoons sugar
2 cups water
6 teaspoons sugar
2 cups water
4 teaspoons sugar
3 cups water
Which of these lemonade glasses is the most concentrated for sugar?
Ос

Oa

Answers

6 teaspoons of sugar

An ambulance is traveling east at 62.4 m/s. Behind it a car travels along the same direction at 34.5 m/s. The ambulance driver hears his siren with a wavelength of 0.47 m. What wavelength would a stationary observer behind the ambulance measure for the sound? The velocity of sound in air is 343 m/s.

Answers

Answer:

The answer is "0.5555 m"

Explanation:

Where the reference leaves the list and the viewer is at rest:

[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]

   [tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]

   [tex]=0.5555 \ m[/tex]

A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy, what is the ratio of the brick's mass to the rock's mass

Answers

Answer:

let M be the mass of brick

let m be the mass of the pebble

.5M(3)^2 =KE

.5m(5)^2= KE

.5M9 = .5m25

9M = 25m

(9/25)M = m

Answer:

A

Explanation:

e=1/2M[tex]V^{2}[/tex]

1/2[tex]M_{1}[/tex]9=1/2[tex]M_{2}[/tex]25

[tex]\frac{M_{1} }{M_{2} }[/tex] =[tex]\frac{25}{9}[/tex]

calculate the aceleration of a vehicle wich start with a zero meter per second, and acelerates to 34 m/s in 21 s.

HELP PLSSS

Answers

Answer:

For the aceleration we have:

Vf = Vo + a * t

Clearing "a":

a = (Vf - Vo) / t

Replacing and resolving:

a = (34 m/s - 0 m/s) / 21 s

a = 34 m/s / 21 s

a = 1,61 m/s^2

The aceleration of the vehicle is 1,61 meters per second squared

Helping me please What can smart modern biosensors be used in heath applications?​

Answers

Answer: Biosensors are modern electronic devices which are made up of biological recognition system and a transducer, for processing of signals and to quantify a particular analyte.

Explanation:

In modern medicine, a Biosensors are used to replace some clinical laboratory investigations of biological fluids with the advantages of being easy to use, rapid and robust as well as offering multianalyte testing capability. They are classified based on their biological recognition elements which includes:

--> enzymatic biosensors

--> immuno biosensors

--> DNA and whole cell biosensors.

They can also be classified according to their signal transduction methods. These includes:

--> electrochemical Biosensors,

--> optical biosensors,

--> thermal and mass-based biosensors;

Some of the HEALTH APPLICATIONS of a modern biosensor includes:

--> Monitoring of glucose levels in diabetic patients which makes use of enzyme based sensors to monitor concentration of glucose in patients.

--> Detection of general metabolic status of bacteria, fungi, yeast, animal or plant cells.

--> Monitoring of cholesterol levels to prevent cardiovascular diseases. Cardiac troponin and. C - Reactive Proteins are biomarkers that are easily detected using biosensors.

--> They are used for Cancer clinical testings

what is the acceleration of a softball if it has a mass of 0.50 kg and hits the catcher’s glove with a force of 25 newtons

Answers

Answer:

a = 50 [m/s²]

Explanation:

This type of problem can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force =25 [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

[tex]a=F/m\\a=25/0.5\\a= 50 [m/s^{2}][/tex]

A typical donation of whole blood is equal to a pint (473 ml). If the average density of blood rhoBLOOD = 1060 kg/m^3, what is the mass of donated blood?

a. 5.01 kg
b. 0.501 lb
c. 0.501 kg
d. 0.501 g
e. 5.01 g

Answers

Answer:

c. 0.501 kg

Explanation:

Given;

volume of the denoted blood, V = 473 ml = 0.000473 m³

density of the denoted blood, ρ = 1060 kg/m³

The mass of donated blood is given as;

mass of the donated blood = density of the denoted blood x volume of the denoted blood

m = 1060 x 0.000473

m = 0.501 kg

Therefore, the mass of donated blood is 0.501 kg


Which is not true about waves?
O Compression and rarefaction are found in longitudinal waves.
O Sound waves are transverse waves.
O Amplitude, wavelength, and troughs all describe parts of a wave.
O Electromagnetic waves are transverse waves.

Answers

Answer:

sound waves are transverse waves

If your parasympathetic nervous system was activated what might you be doing?

A. Relaxing under a tree
B. Running a marathon
C. Skiing
D. Driving home in traffic

Answers

A. Relaxing under a tree.

The parasympathetic nervous system is known as the “rest and digest” system. Activities that require as little movement as possible help the heart rate to slow down. The other activities would be associated with the Sympathetic nervous system which is known as the “fight or flight,” or when your heart rate is up and running.

During the pitching motion, a baseball pitcher exerted an average horizontal force of 90 N against the 0.1 kg baseball while moving it through a horizontal displacement of 2.0 m before release. (1) what was the amount of work performed by the pitcher on the baseball (2) If the velocity at the start of the pitching motion was zero, at release the ball was traveling horizontally at which velocity

Answers

Answer:

(1) 180 J

(2) 60 m/s

Explanation:

(1) From the question,

Amount of work performed by the pitcher on the baseball = Force × distance.

W = F×d............... Equation 1

Given: F = 90 N,  and d = 2.0 m.

Substitute into equation 1

W = 90×2

W = 180 Joules.

(2)

F = ma........... Equation 2

Where a = acceleration of the ball.

a = F/m

Given: F = 90 N, m = 0.1 kg.

therefore,

a = 90/0.1

a = 900 m/s².

Using,

v² = u² + 2as ............ Equation 3

Where u = 0 m/s, a = 900 m/s², s  = 2 m.

substitute into equation 3

v² = 0² + 2×900×2

v² = 3600

v = √3600

v = 60 m/s

Pls Help physics 8th class question ​

Answers

Answer:

angle of reflection and angle of incident is always equal

______ Is the distance and direction of an object's change in position from its starting point.
(5 Points)
Distance
Displacement
Velocity
Rate

Answers

Displacement because of you displace something it is not going to be in its original place for instants if you put a vase on a shelf and you move it it’s is not going to be on the shelf anymore

Answer:

Displacement

Explanation:

I took the test.

in what condition do acceleration and deaccelration

Answers

Im not sure what are you asking about. But in physic/ mechanic usually when the object against gravity, its deceleration but when the object follow the gravity its acceleration

Also usually when the object move to the left side the speed will be negative (-) this is probably because of deceleration? since deceleration = -acceleration.

Can someone please help?

Answers

Answer:

Gold = 19.3g/cm³

Explanation:

Given the following data;

Mass = 38.6g

Volume = 2cm³

To find the density;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

[tex]Density = \frac{mass}{volume}[/tex]

Substituting into the equation, we have;

[tex]Density = \frac{38.6}{2}[/tex]

Density = 19.3g/cm³

Therefore, from the table the material is Gold.

Gold is a chemical element and it is the element 79 on the periodic table. Thus, it has an atomic number of 79. The chemical symbol for Gold is Au, it is chemically classified as a transition metal and as a solid at room temperature.

Generally, the chemical element Gold is known to have the following physical properties;

I. Malleable.

II. Ductile.

III. A good conductor of electricity and heat.

Also, it is a non-toxic chemical element with a beautiful lustrous sheen.

from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa at point B. Friction between the water and the pipe walls is negligible. What is the rate of discharge at point B

Answers

Answer:

1.8 m/s

Explanation:

Here is the complete question

The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.

What is the rate of discharge at point B?

Solution

Using Bernoulli's equation,

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³

P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

P₁ - P₂ - ρg(h₂ - h₁) = 1/2ρ(v₂² - v₁²)

(h₂ - h₁) = 5 m

Substituting the values of the variables, we have

700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)

36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)

- 13000 Pa = 500 kg/m³(v₂² - v₁²)

(v₂² - v₁²) = - 13000 Pa/500 kg/m³

(v₂² - v₁²) = -26 m²/s²

By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B

πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m

d₁²v₁ = d₂²v₂

v₁  = v₂(d₂/d₁)²

= v₂(0.15/0.05)²

= v₂(3)²

= 9v₂

So, substituting v₁ = 9v₂ into the above equation, we have

(v₂² - v₁²) = -26 m²/s²

v₂² - 9v₂² = -26 m²/s²

- 8v₂² = -26 m²/s²

v₂² = -26 m²/s² ÷ (-8)

v₂² = 3.25 m²/s²

taking square root of both sides,

v₂ = √(3.25 m²/s²)

= 1.8 m/s

So, the rate of discharge at point B is 1.8 m/s

6. a. find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm. B. the electron makes a transition from the n

Answers

Complete Question

(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .

(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.

Answer:

A

  [tex]\Delta E = 337 \ eV[/tex]

B

  [tex]\lambda = 3.439 *10^{-9} \ m[/tex]

Explanation:

Considering question a

From the question we are told that

 The width of the box is  [tex]w = 0.125 \ nm = 0.125 *10^{-9} \ m[/tex]

Generally the energy level of a particle confined to a box is mathematically represented as

         [tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]

Generally the excitation energy is mathematically represented as

         [tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]

From the question  [tex]n_2 = 3\ (Third \ excited \ level ) \ \ and \ \ n_1 = 1[/tex]  

   Here h  is the Planck's constant with a value  of  [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

         m is the mass of  electron with value  [tex]m = 9.11 *10^{-31} \ kg[/tex]

So

        [tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ][/tex]

=>     [tex]\Delta E = 539 *10^{-19} \ J[/tex]

=>     [tex]\Delta E = \frac{539 *10^{-19}}{1.60 *10^{-19}} \ J[/tex]

=>     [tex]\Delta E = 337 \ eV[/tex]

Considering question b

Generally the energy level of a particle confined to a box is mathematically represented as

         [tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]

Generally the excitation energy is mathematically represented as

         [tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]

From the question  [tex]n_2 = 4 \ \ and \ \ n_1 = 1[/tex]

 Here h  is the Planck's constant with a value  of  [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

         m is the mass of  electron with value  [tex]m = 9.11 *10^{-31} \ kg[/tex]

So

        [tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ][/tex]

=>      [tex]\Delta E = 578 *10^{-19} \ J[/tex]

=>      [tex]\Delta E = \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}[/tex]  

=>      [tex]\Delta E = 361.45 \ eV[/tex]  

Gnerally the wavelength is mathematically represented as

          [tex]\lambda = \frac{hc}{\Delta E }[/tex]

=>       [tex]\lambda = \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19} }[/tex]

=>       [tex]\lambda = 3.439 *10^{-9} \ m[/tex]

Determine the weight of a 5.1 kg scooter that moves with a constant acceleration of 3.0 ms2. (Make sure you use the weight equation: W or Fg = mg)

Answers

Answer:

15.3 sorry if i got it wrong

Explanation:

A model airplane with mass 1.0 kg is held by a wire so that it flies in a horizontal circle with radius 20.0 m. The airplane engine provides a net thrust of 1.0 N perpendicular to the wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in this horizontal flight.

Answers

Answer:

330

Explanation:

(a) The torque the net thrust produces about the center of the circle is of 20 N-m.

(b) The angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².

Given data:

The mass of model airplane is, m = 1.0 kg.

The radius of horizontal circle is, r = 20.0 m.

The magnitude of net thrust by engine is, F = 1.0 N.

(a)

The effort made to turn any object is known as the torque. The mathematical expression for the torque is given as,

T = F × r

Solving as,

T = 1.0 × 20.0

T = 20 N-m

Thus, we can conclude that the torque the net thrust produces about the center of the circle is of 20 N-m.

(b)

The expression for the angular acceleration of airplane during the horizontal flight is given as,

[tex]T = I \times \alpha[/tex]

Here, I is the moment of inertia of airplane and its value is,

[tex]I = \dfrac{1}{2}mr^{2}\\\\\\I = \dfrac{1}{2} \times 1.0 \times 20^{2}\\\\\\I =200 \;\rm kg.m^{2}[/tex]

So, the angular acceleration is,

20 = 200 × α

α = 20/200

α = 0.1 rad/s²

Thus, we can conclude that the angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².

Learn more about the torque here:

https://brainly.com/question/19247046

with a mass if 74 kg, is a moving with a velocity of 11m/s, what is the kinetic energy ?

Answers

Answer:

4477 J

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 74 \times {11}^{2} \\ = 37 \times 121[/tex]

We have the final answer as

4477 J

Hope this helps you

Which water would you use to make salt dissolve the slowest?

Boiling water
Cold water
Hot water
Room temperature water

Answers

Answer:

boiling water because salt dissolves quicker in hot Water and the hottest is boiling

A block has mass of 20 gm and volume of 4 m3 , calculate the density *

Answers

Answer:

Density = 5 g/m³

Explanation:

Given:

Mass of block = 20 gm

Volume of block = 4 m³

Find:

Density

Computation:

Density = Mass / Volume

Density = 20 / 4

Density = 5 g/m³

5 kg/m3

Explanation:

The formula to find density is (d) = mass (m) ÷ volume (v)

What is true of the bottom layer of the ocean

Answers

Answer:

well i tough it was infinte but its not an the layer is a d         i       c               k  

Explanation:

A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:

Answers

Answer: Induced emf is given by:

ε= Bvl

On putting the values we get

=5×10

−5

×1.50×2

=0.15mV

Explanation:

Hope these helped have a wonderful Christmas break ❄️

A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block is released from rest when the
spring is stretched + 0.15 m as shown in figure below. At the instant the block is released the force vector on the block is.
(Consider Right direction is positive direction and Left direction is negative direction)
- 34.29 N
+ 34.29 N
+ 24 N
Оe
h Oa-241

Answers

Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

Mass of the block = m = 0.7 kg

Sprint constant = k = 160 N / m

x = 0.15m

Now the force on the block is

F = kx

= (160) (0.15)

= 24 N

As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

how to calculate the speed using time and distance

Answers

Answer:

speed = distance/time

Explanation:

distance -> s

speed -> v

time -> t

The calculation of speed using time and distance shows that speed is equal to distance/time.

What is speed?

Speed refers to the rate of movement of a vehicle or person.

Thus, the speed can be computed by dividing the total distance by the time consumed.

Learn more about speed, distance, and time at https://brainly.com/question/4931057

Find the resultant of the following displacement:
A = 20 Km 30° south of east;
B = 50 Km due west;
C = 40 Km north east;
D = 30 Km 60° south of west.

Answers

Answer:

Explanation:

Basically you just have to find the left vectors. To do so divide A, C and D into horizontal and vertical vector. A: 10km to south and 10root3 to east. Just sine and cosine of 30 at 20km. D: 15 km to west and 15root3 to south. Again sine and cosine of 60 at 30 km. C: 45 degrees so 20root2 to north and east. Add all these up with B. Then you have 7.696 km due south and 19.395 km due west. Resultant displacement magnitude = root(7.696^2+19.395^2)=20.866 south of west with angle=arctan(7.696/19.395)=21.644 degrees

If you are holding a stack of books at your waist and then you walk across the room your hands did?

Answers

Answer: 3) work

Explanation:

Work is doing something that requires you to expend energy. In walking across the room while your hands were carrying the stack of books, you expended energy through them which means that they did work.

We can neither tell the efficiency nor the power taken to do the work above from the given information. And by moving and holding books, there was definitely work done which means that all options apart from 3 are false.

Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 17600 N for 55 ms.

Answers

Answer:

0.8 m/s

Explanation:

F(avg) * Δt = I, where

F(avg) = Average force

Δt = change in time

I = impulse

From the question, we know that the

average force is given as -17600 N

time change is given as 55 milliseconds

speed of the player, v = 8 m/s

Mass of the player is given as 110 kg

Impulse, I = F(avg) * Δt

I = -17600 * 0.055

I = -968

Now, using the Impulse-Momentum theory, we have that

Δp = I and thus,

Δp = m [v(f) - v(i)], where

Δp = change in the momentum

v(f) = final speed of the player

v(i) = initial speed of the player

Substituting the values, we have

I = m [v(f) - v(i)]

-968 = m.v(f) - m.v(i)

m.v(f) = m.v(i) - 968

110v(f) = 110 * 8 - 968

110v(f) = 880 - 968

110v(f) = -88

v(f) = -88 / 110

v(f) = -0.8 m/s

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