Heat waves are defined by the difference in temperature compared to the normal _________ typical of a region.

Answer:

Helen Keller was born on June 27, 1880 in the small town of Tuscumbia, Alabama. When she was a year old, she was stricken with an illness that left her without sight or hearing. In the early years after her illness, it was difficult for her to communicate, even with her family; she lived her life entirely in the dark, often angry and frustrated with the fact that no one could understand her. Everything changed in March of 1887, when Helen's teacher, Anne Sullivan, came to live with the family in Alabama and turned Helen's world around.

Miss Sullivan taught Helen the names of objects by giving them to her and then spelling out the letters of their name in her hand. Helen learned to spell these words through imitation, without understanding what she was doing, but eventually had a breakthrough and realized that everything had a name, and that Miss Sullivan was teaching them to her. From this point on, Helen acquired language rapidly; she particularly enjoyed learning out in nature, where she and her teacher would take walks and she would ask questions about her surroundings. Soon after this, Helen learned how to read; Miss Sullivan taught her this by giving her strips of cardboard with raised letters on them, and then having her act out the sentence with objects. Soon, Helen could read entire books.

In May 1888, Helen went north to visit Boston with her mother and teacher. She spent some time studying at the Perkins Institute for the Blind, and quickly befriended the other blind girls who were her age. They spent a vacation at Brewster in Cape Cod, where Helen experienced the ocean for the first time. Following this, they spent nearly every winter up north.

Once she had learned to read, Helen was determined next to learn how to speak. Her teacher and many others believed it would be impossible for her to ever speak normally, but she resolved to reach that point. Miss Sullivan took her to the Horace Mann School in 1890 to begin learning with Miss Sarah Fuller, and Helen learned by feeling the position of Miss Fuller's lips and tongue when she spoke. The moment she spoke her first words, "It is warm," was a powerful memory for her: she was thrilled that she might be able to speak to her family and friends at last.

The winter of 1892 was a troubling time for Helen. Seemingly inspired by the beautiful fall foliage around her, she wrote a story called "The Frost King," and sent it up to her teacher at the Perkins Institute as a gift. It soon came out that Helen's story was quite like another in a published book, called "The Frost Fairies." Helen had been read the original story as a child, and the words had remained so ingrained in her mind that she'd unwittingly plagiarized them when she wrote her own story. This tainted Helen's relationship with her Perkins Institute teacher, Mr. Anagnos, and made her distrust her own mind and the originality of her thoughts for a long time.

In 1894 Helen attended the Wright-Humanson School for the Deaf in New York City, and began studying formal subjects like history, Latin, French, German, and arithmetic. In 1896, she began her studies at the Cambridge School for Young Ladies in Massachusetts, which would prepare her to eventually attend Radcliffe College, the women's college affiliated with Harvard University. This was her first time attending school with girls who could see or hear, rather than other students who were also deaf or blind. Though it was a challenge, she persevered; however, her mother eventually withdrew her from the Cambridge School to finish her Radcliffe preparation with a private tutor, because they did not agree with the Cambridge School principal's wish to lighten Helen's course load. She successfully qualified for Radcliffe in 1899, and entered college in the fall of 1900. Though college presented unique obstacles for Helen to overcome, she deeply appreciated her opportunity to attend.

Helen uses the final chapters of her memoir to discuss certain things that are particularly important to her, like her love of books, her favorite pastimes, and the friends she made who shaped her life. Two additional sections of the autobiography include Helen's personal letters written throughout her youth, as well as supplementary commentary by her editor, with a first-hand account by 

or

The Story of My Life (1903) chronicles the early years of Helen Keller, a young woman who became both deaf and blind at a young age. The book explores the challenges she faced growing up as a child with disabilities, and introduces the amazing people who helped her along the way.

Explanation:

here

energy at first = mg (3) J = 29.4J

Energy after bouncing: mg(2) = 19.6 J

Energy changed by the ball: 29.4J - 19.6J = 9.8J

Heat energy or sound energy are two ways that this energy is lost.

Work performed during the bounce equals 9.8J.

d. Momentum change = mvf-mvi

29.4 = 1/2mvi2

vf =7.668m/s

19.6J =1/2mvf2

vf=6.261m/s

Change of momentum is equal to 1kg x (-1.407m/s)/1kg = -1.407 kg/s (m(vf-vi)).

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When two small ball collide elastically in which first ball is moving with velocity, v with second ball which is in rest than after collision the second ball will move with velocity, v and first ball will come to rest.

An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.

An elastic collision is one in which the momentum, p and the kinetic energy, KE are both conserved.

[tex]p_{i} =p_{f}[/tex]

[tex]KE_{i} = KE_{f}[/tex]

Given in the question two identical steel ball that means have same mass and one is moving with velocity, v another is in the rest initially, after collision the momentum of first ball will transfer to second ball so second ball will move with velocity, v and first ball will come to rest.

So final velocity of first ball is zero and second ball is v.

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A. The speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is 3.65m/s

B. The acceleration when the elevator is 1.00 {\rm m} below point where it first contacts a spring is 4m/s²

In calculating the speed of the elevator and acceleration, first we have to find the force of gravity F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:

F = mg

Where:

Mass of the elevator; m= 2000kg

Acceleration due to gravity; g = 9.8m/s

2000kg × 9.8m/s²= 19600N

F = 19600

Force of opposing friction clamp of gravity = 17000N

Net force on the elevator = force of gravity - Force of opposing friction clamp

Net force on the elevator = 19600 - 17000

Net force on the elevator = 2600 N

We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:

K.E = 1/2 mv²

Where

Mass of the elevator; m = 2000kg

Velocity of the elevator = 4.00m/s

K.E = (1/2)*2000kg*(4m/s)²

K.E = 16000J

The kinetic energy and energy gained will be absorbed by the spring across the next 2m

Therefore,

Energy; E = K.E + P.E

Where:

Kinetic energy K.E = 16000J

Potential Energy P.E = ?

P.E of spring = net force absorbed × distance at compression

Where:

Net force absorbed = 2600N

Distance at compression = 2.0m

P.E = 2600*2

P.E = 5200J

E = 16000J + 5200J

E = 21200J

Spring constant = k

To find k

Using:

E = (1/2)*k*(x)²

Where:

E = 21200J

k = ?

x = 2m

21200J = (1/2)*k*(2m)²

21200J*2 = (4m)k

K = 42400J/4m

K = 10600N/m

Therefore,

Acceleration at 1m compression = ?

Using:

F = K*X

Where

F is force provided by the spring = 10600N/m,

K = 10600 N/m

X = 1m

F = 10600N/m * 1m

F = 10600N (upward)

A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?

Using:

Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy

16000N + 2600N = (1/2)mv² + (1/2)k x²

18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)

18600 = 1000(v²) + 5300

18600 - 5300 = 1000(v²)

13300 = 1000(v²)

V² = 13.300

V =3.65m/s

B. The acceleration of the elevator is 1.00m below point where it first contacts a spring

Spring constant = net force on the elevator + resultant force

Where:

Spring constant = 10600N

Net force on the elevator = 2600N

Resultant force = ?

10600N = 2600N + resultant force

Resultant force = 10600N - 2600N

Resultant force = 8000N

Using the equation for Newton's 2nd law where F = ma,

a = F/m

Where:

Resultant force; F =8000N

Mass of the elevator; m =2000kg)

a = 8000 / 2000

a = 4m/s²

Here's the complete question:

In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.

1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?

2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?

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The force exerted by the parachutist on the ground is 471.24 N when she comes to rest at a distance of 0.660 m

Let t = Time taken

u = Initial velocity , v = Final velocity

s = Displacement, a = Acceleration

As given in the problem initial velocity u = 3.85 m/s

Final velocity v = 0 (rest)

s = 0.660 m

From the equation of motion v² = u² + 2as

we have a = v²-u² / 2s

⇒a = 0-3.85² / (2* 0.660)

⇒a = -11.22 m/s²    (negative sign shows deceleration)

we know force F = mass m * acceleration a

Given mass of parachutist m = 42 kg

∴Force exerted by parachutist on ground

F = 42 * 11.22 = 471.24 N

If the distance is shorter, acceleration increases in magnitude and so does the force exerted by the parachutist.

The question is incomplete and the full question is probably " A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant acceleration over a distance of 0.660 m, what force does the ground exert on her? "

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In 1912 Wertheimer discovered the phi phenomenon, an optical illusion in which stationary objects shown in rapid succession, transcending the threshold at which they can be perceived separately, appear to move.

When two neighbouring optical stimuli are presented in alternation with a relatively high frequency, an apparent motion is seen that is known as the phi phenomenon. At higher frequencies, beta movement is visible, but the stimuli themselves don't seem to move.

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Answer:

ans a

Explanation:

It is not safe to drive at this speed since the centrifugal force is larger than the force due to friction.

The centrifugal force on car is

F=mv²/r

F= 1600*27*27/150

F=7776N

force due to static friction is

Fst=μN

   = 0.4*1600*9.8

   =6272N

The frictional force between a washer and a metal rod is then compared to the centrifugal force necessary to overcome static friction between a washer and a rod that is experiencing rotational acceleration. This equivalence allows us to compute the theoretical velocity at which the washer starts to slip off the rod. Static friction now enters the picture to prevent slipping before it even begins. By acting radially inward to the turn, it resists the motion by acting in the opposite direction. As a result, if you think of the turn as a portion of a circle, static friction will act as the centripetal force.

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The block will incline up with projectile motion until 0.085 m before sliding down.

We need to know about the projectile motion to solve this problem. The projectile motion is known as parabolic motion and the velocity is divided by 2 axes.

vox = vo cosA

voy = vo sinA

where vox is initial velocity of x axis, voy is initial velocity of y axis, vo is initial velocity and A is the angle.

From the question above, the given parameters are

vo = 5 m/s

A = 20⁰

By using the uniform motion, where vty = 0 m/s. Hence,

vty² = voy² - 2gh

0 = 5.sinA - 2 . 10 . h

20h = 5 . sin20⁰

h = 1/4 . 0.34

h = 0.085 m

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a)current through a battery I=ΔVJR b) emf of the battery using is e=I(r+R).

c) emf in terms of r. R and ΔV is e=ΔV/R(R+r).

Electricity, which is the flow of electrons, or current, over a wire, creates electric and magnetic fields, which are imperceptible zones of energy (also known as radiation).

Voltage is the force used to force the electrons through the wire, much like water is pushed through a pipe, creating an electric field. The electric field gets stronger as the voltage rises. Volts per metre (V/m) is a unit of measurement for electric fields. EMF, or electromagnetic fields, are the collective term for electric and magnetic fields. Radiation from electromagnetic sources produces the electric and magnetic forces found in EMFs. EMFs can be divided into two primary groups:

X-rays and gamma rays are examples of higher-frequency EMFs. These electromagnetic fields (EMFs) fall within the category of ionising radiation.

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Answer:

Move at the same velocity

Explanation:

According to newton's 1st law of motion, an object at rest remains at rest or if in motion, it will remain in motion at constant velocity unless acted upon by a net external force

The new voltage value of parallel-plate capacitor is 800 V.

We need to know about the capacitance of the parallel-plate capacitor to solve this problem. The capacitance of the parallel-plate capacitor depends on the charge and voltage. It can be written as

C = Q/V

where C is capacitance, Q is charge and V is voltage.

Capacitance also can be calculated by the area and distance between capacitor plate

C = εo . A / d

where εo is vacuum permittivity (8.85 x 10¯¹² F/m), A is area of plate and d is distance.

From the question above, we know that

V1 = 200 V

d1 = d

d2 = 4 d

By substituting the given parameters, we can calculate the voltage

εo . A / d = Q/V

The voltage is proportional to the distance of plate. Hence, the ratio of voltage will be

V1 / V2 = d1 / d2

200 / V2 = d / 4d

V2 = 4 x 200

V = 800 N

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The magnitude and direction of Clare's resultant displacement is 26.4 m and  37.3⁰ respectively.

The magnitude of Clare's resultant displacement is calculated as follows;

her initial position, a = 16 m north

Final position, b = 21 m east

Resultant displacement, d = √a² + b²

d = √(16² + 21²)

d = 26.4 m

The direction of the resultant displacement is calculated as follows;

tan θ = 16 / 21

where;

tan θ = 0.762

θ  = arc tan(0.762)

θ  = 37.3⁰

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Answer:

W' = 287.55 N

Explanation:

We know that weight force is defined as follows:

W = m × g

So:

On Earth (W = 1775 N):

1775 = m × 9.8

m = 1775 / 9.8

m = 181.12 kg

On the Moon (g' = 1.62 m/s²):

W' = m × g'

W' = 181.12 × 1.62

W' = 293.42 N

W is the work done by charge from A to B and from C to B then the potential difference  (vc−va)=552.62V.

The difference between the energy levels that charge carriers have at two places in a circuit is known as the potential difference.

Voltage, commonly known as potential difference (p.d. ), is a unit of electrical potential. The charge carriers in a circuit pass through the electrical components, transferring energy to them. We gauge the potential difference using a voltmeter (or voltage).

V = I x, the potential difference formula

The quantity of current times the resistance equals the potential difference (also known as voltage), which is equal to that. When a charge travels between two locations in a circuit, a potential difference of one Volt is equivalent to one Joule of energy being used by that charge.

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Flow rate and velocity are physically connected but very different quantities. The flow rate of the river increases with increasing water velocity. However, river size also affects flow rate.

[v*π*(2/2)^2] *60 = 20*1000 [1 L = 1000 cm^3]

(22*60/7)*v = 20000,

or,  v = 7*20000/22/60or  

v = 106.06 cm/s = 1.06 m/s.

Hence, velocity of the water when its in the hose is 1.06 m/s.

Consider the rate at which a river flows to understand the difference. The flow rate of the river increases with increasing water velocity. However, river size also affects flow rate. For instance, the Amazon River in Brazil carries much more water than a swift alpine stream. The exact formula for the link between flow rate Q and velocity is

A is the cross-sectional area, and

V is the average velocity, therefore Q = A v.

This equation makes sense to me. According to the relationship, the size of a river, pipe, or other conduit, as well as the magnitude of the average velocity (hereinafter referred to as the speed), directly affect the flow rate. The cross-sectional area of a conduit increases with its size.

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Answer:

it will decrease

Explanation:

because the ice will melt

Polar jet streams are formed because of the warm and cold air masses.

Polar front jet stream, also called polar front jet or mid-latitude jet stream stream, a belt of powerful upper level winds that sits at top the polar front.

The winds are strongest in the troposphere, which is the upper boundary of the troposphere, and move in a generally westerly direction in mid-latitudes.

The vertical wind shear which extends below the core of this jet stream is associated with horizontal temperature gradients that extend to the surface. As a consequence, this jet manifests itself as a front that marks the division between colder air over a deep layer and warmer air over a deep layer.

Polar jet streams are formed due to warm and cold air masses.

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Answer:

C

Explanation:

Cora rotational speed is the same as Cameron's rotational speed.

merry-go-angular round's velocity is 2n = 2x 6.06/60 =.1x radian/s

This will represent the system's overall rotating speed, including Cora and Cameron's. It is independent of where they are in relation to the merry-go-center. round's

Cora's rotating speed is equal to Cameron's rotational speed.

The number of rotations a spinning system completes within a specified amount of time can be used to measure rotational speed. Pump speed is often expressed in min-1, whereas rotational speed is measured in s-1 (rev/s) (rpm).

Therefore, the rotational speed of a pump is determined by the pump shaft's spinning frequency (n). It is never specified as a negative number and is not to be confused with particular speed (ns).The impeller's designated direction of rotation—clockwise when moving to the right with regard to the direction of inflow—is distinct from the pump's required clockwise or anti-clockwise rotation.

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Answer: They are bigger

Explanation: They are the biggest planets in our Solar System and attract more to them so they would obviously have more moons than all of the other planets.

Answer:

Explanation:

Because of the huge mass of these planets.

The change in gpe ( gravitational potential energy ) of trophy of mass 3 kg that is lifted 2 m up in the air is 58.8 J

ΔPE = m g h

ΔPE = Change in potential energy

m = Mass

g = Acceleration due to gravity

h = Height

m = 3 kg

h = 2 m

g = 9.8 m / s²

ΔPE = 3 * 9.8 * 2

ΔPE = 58.8 J

Gravitational potential energy of an object is due to the presence of gravity. The higher the object is from the point of surface from which the gravity is exerted, higher will be its potential energy.

The change in gpe of the trophy is 58.8 J

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The acceleration when a 44.7 N net force is applied to 7.7 kg object is 5.8 m / s²

According to Newton's second law of motion,

F = m a

F = Force

m = Mass

a = Acceleration

m = 7.7 kg

F = 44.7 N

a = F / m

a = 44.7 / 7.7

a = 5.8 m / s²

Newton's second law of motion states that the force is equal to the rate of change of momentum. It can also be said as acceleration of an object is directly proportional to the net force acting on the object.

Therefore, the acceleration of the object is 5.8 m / s²

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The friction that is necessary to keep her from falling off the merry go round is 280.16N, the distance travelled in one revolution is 28.28 meters and the time period T after in revolution is 6.89 seconds.

The mass of the worker is 75 kg, the radius of merry go round is 4.5m and the linear velocity of the merry go round is 4.1m.s.

a) The diagram is attached below.

b) The friction force while rotation will be equal to the centrifugal force, this is the reason why he will be able to go round,

So, we can write,

Friction force Fr = Centrifugal force

Fr = MV²/R

Where Fr is the friction force,

M is the mass of the worker,

V is the velocity of the merry go round,

r is the distance at which the worker is standing.

Putting all the values,

Fr = 75 x 4.1 x 4.1/4.5

Fr = 280.16N.

The friction required is 280.16N.

c) The distance she travelled by her after the first round will be equal to the circumference of the circle,

Distance = 2Πr

D = 2 x 22 x 4.5/7

D = 28.28m

d) time period T after one revolution,

T = Distance/velocity

T = 28.28/4.1

T = 6.89 seconds.

The time period is 6.89 seconds.

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Answer:

F = m a = K M m / R^2         force on mass on moon

a = K M / R^2

a = 6.67E-11 * 7.36E22 / (1.74E6)^2

a = 6.67 * 7.36 / 1.74^2 * 10^-1 = 1.62 m/s^2      about 1/6 of earth

Answer:

c

Explanation:

because whenever you push g box into h box and then h displaces according to third law of motion h box will apply the force on g box so that's why

There were 1647 before liftoff. 38 m moving planes.

An object is considered to be in motion if its location throughout time changes in relation to its surroundings. It is the gradual change in an object's location. The only kind of motion that exists is linear motion.

The three equations are as follows: S = ut + 12at2, v = u + at, v2 = u + 2as.

The airplane's initial speed is U=0 m/s.

The plane accelerates at a 4 m/s rate.

with a time stamp of 28.7 seconds.

Distance traveled using the formula S=ut+a

⇒ S=0(28.7)+0.5(4) (4)

=1647.38 ~ 1647m

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The impulse of the two skaters is the same but the speed of Jason is only half that of Megan.

We define the impulse as the product of the force and the time. We know that two objects are taking of at the same time. Given the fact the they are taking off from the same kind of surface at the same time, the impulse of the two are the same.

We know that the speed of the object would depend on the mass of the object. The greater the mass, the lesser the speed of the object hence Jason would have a lesser speed than Mason.

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It takes 2.916s for the bicyclist to catch the runner and the runner had ran 17.2m with constant velocity of 3.5m/s

How quickly or slowly an object is moving can be determined by its velocity and speed. The need to determine which of two or more moving objects is moving faster arises frequently in our daily lives.

If both vehicles are traveling down the same road in the same direction, it is simple to determine which is moving more quickly.

To tell who is moving faster, though, is challenging if they are moving in the opposite direction. The idea of velocity is valuable in such circumstances

Lets say runner and bicyclist start at a point vi

The distance bicyclist cover to catch runner  is [tex]S = (v_i)t = \frac{1}{2} at^2[/tex]

As the initial velocity is zero distance [tex]S = \frac{1}{2} at^2[/tex]

For the runner, distance he could run before bicyclist overpassed him is

S = vt

Now as athletes was 7m ahead so

= (athlete)S  = (bicyclist)S

= (athlete)S = (bicyclist)S - 7

Substitute the values

= vt = [tex]\frac{1}{2} at^2[/tex]

= 3.5t = [tex]\frac{1}{2} (2.4)t^2[/tex]

= 3.5t = 1.2t²

= 3.5 = 1.2t

= t = 2.916s

Distance of runner = 3.5 × 2.916s

                                = 10.206 + 7m

                                = 17.2m

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The action of volleying a ball straight up and down comes under the free-fall motion of the ball. Free-fall is defined as the motion of a body which falls freely due to the gravitational pull of the earth.

The velocity (v) of the ball undergoing free-fall can be calculated using the free-fall formula,

[tex]v^{2}[/tex] = 2gh

where, g is the acceleration due to gravity and h is the height of the freely falling body.

Given that, the height to which ball rises, h = 3.6m. (g = 9.8 m/[tex]s^{2}[/tex])

Substituting h and g in the velocity formula,

 [tex]v^{2}[/tex] = 2 × 9.8 m/[tex]s^{2}[/tex] × 3.6m

 [tex]v^{2}[/tex] = 70.56 [tex]m^{2}/s^{2}[/tex]

The velocity of volley ball is,

v =  [tex]\sqrt{70.56m^{2}/s^{2} }[/tex]

∴ v = 8.4 m/s

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