what is the distance between a(-6,3),b(-2,4),c(-8,3)​

Answer:

In the present study, the startle blink reflex is used as a measure of emotion regulation to effective picture stimuli. Based on the aphasic theory of emotion, it is hypothesized that the startle response will be largest in magnitude in the presence of negative emotional stimuli (Varanasi, Spence, & Lang, 1988).

If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

Given the data in the question;

Since the hockey puck was initially in the referee's hands

First we will find the velocity of the Puck when it hits the ground

From the Third Equation of Motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

[tex]v^2 = u^2 + 2gh[/tex]

We substitute in our value and find "v"

[tex]v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s[/tex]

Now, Velocity of the hock puck after it hits the ground and bounce back;

We know that; Coefficient of restitution [tex]= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}[/tex]

Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision

we substitute in our values;

Relative Velocity after collision [tex]= 0.35 \ *\ 6.85857m/s[/tex]

Relative Velocity after collision [tex]= 2.4 m/s[/tex]

Now, to determine how high should the puck bounced back

We use the Third Equation of Motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

[tex]v^2 = u^2 + 2gh[/tex]

Now, since the hockey puck bounces back, it is experiencing a negative acceleration

Hence, the equation becomes

[tex]v^2 = u^2 - 2gh[/tex]

We substitute our values into the equation and find "h"

[tex](0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m[/tex]

Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

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Answer:

include the statements pls so i can choose wich one it is and tell you

Explanation:

Focus on the ball. Do you see you? Keep following it. Put all your energy into watching it and then unlock your third eye. With that you'll be able to determine your question.

Answer:

60°

Explanation:

It will be equilateral triangle, you can prove with the cosine theorem.

1)P^2=P^2+P^2 - 2×P×P× cos A

2) cos A= P^2 / 2× P^2 = 1/2

cos A = 1/2

A = 60°

If you wrote this question right,it will be solved like this :)

Answer:

b

Explanation:

hope helpful

have a great day

Answer:

False

Explanation:

Answer:

The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.

Explanation:

The force that will move the box at constant velocity must be a little more than 64 N.

The coefficient of sliding friction is obtained from the formula;

μ= F/R

Where;

F = frictional force

μ = coefficient of sliding friction

R = Normal reaction

It is necessary to note that the force that will move the body must be greater than the frictional force acting between the body and the surface in order to move the body. Hence, the force that will move the box at constant velocity must be a little more than 64 N.

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In order to calculate frictional force look below..

The formula given by

[tex]\\ \sf\longmapsto F_f=\mu N[/tex]

Or

[tex]\\ \sf\longmapsto F_f=\mu mg[/tex]

u is coefficient of friction

N is normal reaction.

this is the correct answer

Answer:

c

Explanation:

Answer:

A syncline is visible.

Explanation:

A syncline is visible.

Answer:

These mountains show no evidence of folding.

Explanation: Solution

1.

Gravitational potential energy

U=mgh=500*9.8*50

U=245000 J

2.

Kinetic energy is present at bottom of the hill

K=(1/2)mV2=(1/2)*500*27.82

K=193210 J

3.

Work done by friction

W=193210-245000=-51790 J

The mechanical energy at the top and bottom of the hill is equal to 352800 J and 163840 J respectively.

Kinetic energy (KE) can be described as the energy possessed by a moving object due to its motion. Work by a body will be done to change the kinetic energy. The kinetic energy is represented as K.E = ½mv².

Potential energy (P.E) can be described as the energy that is stored by an object due to its position and is represented in the equation as P.E = mgh, where ‘m’ is the mass, ‘g’ is the acceleration due to gravity and ‘h’ is the height.

The mechanical energy = Kinetic energy + potential energy

Given, the mass of the car, m = 500 Kg

The height of the hill, h = 72 m

The velocity of the car, v = 25.6 m/s

At the top of the hill, the mechanical energy = potential energy

The potential energy at the top of the hill, = mgh

P. E. = 500 × 9.8 ×72

P.E. = 352800 J

At the bottom of the hill, the mechanical energy = kinetic energy

The kinetic energy of the car at the bottom of the hill,

K.E. =  ½ × 500 (25.6)²

K.E. = 163840 J

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#SPJ2

Answer:

the correct answer is sugar

Answer:

Explanation:

F = ma

Assuming the 20° is angle θ measured to the horizontal

mgsinθ - μmgcosθ = ma

g(sinθ - μcosθ) = a

at constant velocity, a = 0

g(sinθ - μcosθ) = 0

   sinθ - μcosθ = 0

                 sinθ = μcosθ

μ = sinθ/cosθ

μ = tanθ

μ = tan20

μ = 0.3639702342...

μ = 0.36

Answer:

bowling ball, basket ball, tennis ball

Explanation:

Answer:

A

Explanation:

all galaxies exploded in order to create the sun/stars

Answer:

What is B physical manipulation

Explanation:

Physical manipulation means fertilizers that are manufactured, blended, or mixed, or animal manures or compost that have been changed from their initial physical state by manipulations such as drying, cooking, chopping, grinding, shredding, ashing, or pelleting.

Please allow me to know if my answer helped you with a thank you!

Miss Hawaii

The coefficient of kinetic friction for the rough horizontal surface is 0.66.

The given parameters;

Apply work-energy theorem; the work done the force of friction is equal to the energy stored in the spring.

[tex]F_kd_1 = \frac{1}{2} kd_2^2\\\\\mu_kmg cos(\theta)d_1 = \frac{1}{2} kd_2^2\\\\\mu_k(1 \times 9.8 \times cos(30)\times 4) = \frac{1}{2} \times 26.5 \times (1.3)^2\\\\33.95\mu_k = 22.39\\\\\mu_k = \frac{22.39}{33.95} \\\\\mu_k = 0.66[/tex]

Thus, the coefficient of kinetic friction for the rough horizontal surface is 0.66.

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Answer:

mę břöőda I dunno this answer

Explanation:

hope you find it

Explanation:

nnnnkgcbchg of bbjbjk oh jjzjjzkedkkdkfjjsjdsjfkjfdkkdowkronqfojofj see j FC

Answer:

Explanation:

Your question is quite confusing, particularly the information about microsecond = 0.5.

I'm going to ASSUME that you mean coefficient of static friction μs = 0.5

unfortunately typing a subscript "s" is very difficult and probably leads to such confusion.

I will also ASSUME that the block, and spring, and force vector are all horizontal.

If the force is slowly increased until the block slips, the spring will compress until the force on each end equals the maximum static friction force. As we are not concerned with the compression distance, only the force, we can ignore the spring constant information and simply find the maximum available static friction force.

F = μN

F = μmg

F = 0.5(4.0)(9.8)

F = 19.6 N

Not that it matters, but the spring will have extended or compressed 19.6/0.8 = 24.5 m, which is a very long and very light spring

Answer:

0 because she went to exact same place as before

Answer:

reinforcing stimulus

Explanation:

Think its correct

Answer:

reinforcing stimulus

Explanation:

because it increases the occurence of a response

Enzymes. ... This is because heat energy causes more collisions, with more energy, between the enzyme molecules and other molecules. However, if the temperature gets too high, the enzyme is denatured and stops working. A common error in exams is to write that enzymes are killed at high temperatures.

An Excerpt from “Optimism”

by Helen Keller

1 Could we choose our environment, and were desire in human undertakings synonymous with

endowment, all men would, I suppose, be optimists. Certainly most of us regard happiness as

the proper end of all earthly enterprise. The will to be happy animates alike the philosopher, the

prince and the chimney-sweep. No matter how dull, or how mean, or how wise a man is, he feels

that happiness is his indisputable right.

2 It is curious to observe what different ideals of happiness people cherish, and in what singular

places they look for this well-spring of their life. Many look for it in the hoarding of riches, some

in the pride of power, and others in the achievements of art and literature; a few seek it in the

exploration of their own minds, or in the search for knowledge.

3 Most people measure their happiness in terms of physical pleasure and material possession.

Could they win some visible goal which they have set on the horizon, how happy they would be!

Lacking this gift or that circumstance, they would be miserable. If happiness is to be so

measured, I who cannot hear or see have every reason to sit in a corner with folded hands and

weep. If I am happy in spite of my deprivations, if my happiness is so deep that it is a faith, so

thoughtful that it becomes a philosophy of life,—if, in short, I am an optimist, my testimony to

the creed of optimism is worth hearing....

4 Once I knew the depth where no hope was, and darkness lay on the face of all things. Then

love came and set my soul free. Once I knew only darkness and stillness. Now I know hope and

joy. Once I fretted and beat myself against the wall that shut me in. Now I rejoice in the

consciousness that I can think, act and attain heaven. My life was without past or future; death,

the pessimist would say, “a consummation devoutly to be wished.” But a little word from the

fingers of another fell into my hand that clutched at emptiness, and my heart leaped to the

rapture of living. Night fled before the day of thought, and love and joy and hope came up in a

passion of obedience to knowledge. Can anyone who has escaped such captivity, who has felt

the thrill and glory of freedom, be a pessimist?

5 My early experience was thus a leap from bad to good. If I tried, I could not check the

momentum of my first leap out of the dark; to move breast forward is a habit learned suddenly

at that first moment of release and rush into the light. With the first word I used intelligently, I

learned to live, to think, to hope. Darkness cannot shut me in again. I have had a glimpse of the

shore, and can now live by the hope of reaching it.

6 So my optimism is no mild and unreasoning satisfaction. A poet once said I must be happy

because I did not see the bare, cold present, but lived in a beautiful dream. I do live in a

beautiful dream; but that dream is the actual, the present,—not cold, but warm; not bare, but

furnished with a thousand blessings. The very evil which the poet supposed would be a cruel

6) Read the last sentence from the text.

Only by contact with evil could I have learned to feel by contrast the beauty of truth and love and goodness.

Explain how Helen Keller develops this idea in the text. Use specific details to

support your answer.

The angular acceleration of the larger camshaft is 995.72 rad/s².

The given parameters;

The angular acceleration of the 3.75 cm camshaft is calculated as follows;

[tex]\omega _f = \omega _i + \alpha t\\\\\omega _f =0 + \alpha t\\\\\omega _f = \alpha t\\\\(1250 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1\min}{60 \ s} ) = 0.032 \alpha \\\\130.92 = 0.032\alpha \\\\\alpha = \frac{130.92}{0.032} = 4091.25 \ rad/s^2[/tex]

The angular momentum of the camshaft is calculated as follows;

[tex]I_1 \alpha _1 = I_2 \alpha_2 \\\\\frac{1}{2} mr_1^2 \alpha _1 = \frac{1}{2}m R^2 \alpha_2\\\\r_1^2 \alpha _1 = R^2 \alpha_2\\\\\alpha_2 = \frac{r_1^2 \alpha _1 }{R^2} \\\\\alpha_2 =\frac{(0.037)^2 \times (4091.25)}{(0.075)^2} \\\\\alpha _2 = 995.72 \ rad/s^2[/tex]

Thus, the angular acceleration of the larger camshaft is 995.72 rad/s².

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