How are stage and film versions of a drama similar the viewer must infer the mood from the lighting and sound effects?

Answers

Answer 1

Stage and film versions of a drama are similar in that the audience must use context clues such as lighting, sound effects, and dialogue to infer the mood of a scene. This requires the audience to pay attention to the details of the performance, as subtle changes in these elements can drastically change the overall tone of the production.

The main difference between a stage and film version of a drama is that a stage production relies heavily on the physicality of the actors and their interactions with one another, while a film production relies heavily on the technical aspects such as camera angles, lighting, and sound design. Additionally, a stage production is performed live in front of an audience, while a film is created in a studio and viewed later by an audience.

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Related Questions

A 15,000 kg rocket traveling at +230 m/s turns on its engines. Over a 6. 0 s period it burns 1,000 kg of fuel. An observer on the ground measures the velocity of the expelled gases to be −1,200 m/s

Answers

First, we need to calculate force of the rocket.

What is Velocity?

Velocity is a vector quantity that describes the rate and direction of a body's motion. It is represented by an arrow pointing in the direction of motion and having a length equal to the rate of motion. Velocity is a fundamental concept in classical mechanics and is defined as the rate of change of position with respect to time.

Calculate the force of the rocket:

F = ma = (15,000 kg)(230 m/s) = 3,450,000 kg m/s2

Calculate the force of the exhaust gases:

F = ma = (1,000 kg)(-1,200 m/s) = -1,200,000 kg m/s2

Calculate the net force on the rocket:

Net Force = F_rocket + F_exhaust = 3,450,000 kg m/s2 + (-1,200,000 kg m/s2) = 2,250,000 kg m/s2

Calculate the acceleration of the rocket:

a = F/m = (2,250,000 kg m/s2)/(15,000 kg) = 150 m/s2

Calculate the change in velocity of the rocket:

Δv = a × t = (150 m/s2)(6.0 s) = 900 m/s.

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You bought a can of apricot jam. The contents marked as: net weight: 500g and volume: 400ml is the statement correct?why? and what is the mass of the jam you bought? calculate the density of jam in g/cm

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The statement is not correct the mass of the jam is 500g and the density of the jam is 0.125gm/cm³

How to calculate the density?The statement is not correct because mass and volume cannot be used together to describe the contents of a can. Mass is a measure of how much matter is in an object and volume is a measure of the amount of space taken up by an object. So, in this case, the can of apricot jam should just be marked with the net weight in grams.The mass of the jam you bought is 500g.To calculate the density of the jam, we need to divide the mass by the volume. The density of the jam is 500g/400ml = 1.25g/ml. To put this in g/cm3, we need to divide the density by 1000, which gives us 1.25g/ml = 0.125g/cm3.

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You throw a ball vertically upward so that it leaves the ground with velocity 5. 68 m/s.


(a) What is its velocity when it reaches its maximum altitude? magnitude 0 Correct: Your answer is correct. M/s direction Incorrect: Your answer is incorrect.

(b) What is its acceleration at this point? magnitude -9. 8 Incorrect: Your answer is incorrect. The response you submitted has the wrong sign. M/s2 direction Incorrect: Your answer is incorrect.

(c) What is the velocity with which it returns to ground level? magnitude -5. 68 Incorrect: Your answer is incorrect. The response you submitted has the wrong sign. M/s direction Correct: Your answer is correct.

(d) What is its acceleration at this point? magnitude 9. 8 Correct: Your answer is correct. M/s2 direction Correct: Your answer is correct

Answers

The length of time needed to descend back to the earth's surface. Gravitational acceleration is equal to -9.8 m/s2. 2 s^2 s2

Explanation for the above answer:

When anything is sent vertically upward, it slows down due to gravity. Up until it reaches a maximum height, where the velocity is zero, its speed is decreasing. After that, gravity accelerates it uniformly downward.

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

What is the vertical velocity formula?

With this knowledge, the vertical velocity formula, vf=vi+gt v f = v I + g t, can be used to calculate the object's terminal velocity. This equation is used by physicists to predict how any object will move on a vertical plane.

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If you cool down a container of gas, there will be _______ collisions with the side of the container, resulting in ______ pressure.

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A container of gas might lose pressure by cooling down because there are fewer collisions with the container's side.

What transpires when a container of gas is cooled?

When the gas in the supplied sealed jar is chilled, the molecules travel more slowly. This occurs because when the temperature drops, gaseous particles get closer to one another (on cooling). Upon cooling, there are also fewer accidents.

How does a gas's compression affect the frequency of collisions inside a container?

The container's capacity has shrunk, which reduces the distance the gas molecules must travel before colliding. As a result, there will be more collisions per second, increasing the pressure.

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Why are dynamic flexibility tests not used as often as?

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Dynamic flexibility tests are not used as often as static flexibility tests because dynamic flexibility tests involve more subjective measurements.

Static flexibility tests are more common in comparison to dynamic tests because static flexibility tests are easier to replicate. It is difficult to standardize motion-based tests such as dynamic flexibility tests between various subjects, while static flexibility tests are simple and easy to standardize it is because the metrics are fixed and not open to interpretation.

Since, dynamic flexibility tests are hard to perform because they measure the amount of stress on a muscle that is in motion. And it is difficult to assess this among multiple people and to get an accurate picture. That is why these tests are not used as often as static flexibility tests.

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Complete question

Why are dynamic flexibility tests not used as often as static flexibility tests?

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Which of the following is the law of action-reaction?
OA. Newton's first law
B. Newton's second lav
OC. Newton's third law
OD. Newton's fourth law

Answers

Answer:

OC.Newton's third law of motion which states that,"every action has equal and opposite reaction."

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A student is pushing on a heavy door, trying to slide it open. Their friend stands behind them and helps push. How have the forces changed

Answers

There will be an increase in the net force.It appears that an object is moving faster, slower, or in a different direction.

What actually constitutes force?

There are several instances of forces in daily life, including:heft and powerthe force a bat applies to a ball.the pressure that a hair brush applies to hair when brushing it.the pressure your foot applies to the pedal when you're riding a bike.

What occurs when a force shifts?

An object can change its size or shape, begin moving, stop moving, accelerate, or decelerate as a result of a pressure applied on it.

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A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its equilibrium position, what is its instantaneous velocity?
A.Instantaneous velocity cannot be determined without additional information
B.At equilibrium, its instantaneous velocity is less than its maximum but not zero.
C.At equilibrium, its instantaneous velocity is at maximum
D.At equilibrium, its instantaneous velocity is zero.

Answers

At the instant when a mass is at its equilibrium position when undergoing simple harmonic motion, its instantaneous velocity is its maximum.

Therefore the answer is C. At equilibrium, its instantaneous velocity is at maximum

The simple harmonic motion is periodic motion that oscillates around an equilibrium position. At the equilibrium position, the net force on the mass is zero. Therefore, the mass is not accelerating, so its instantaneous acceleration is zero.

The motion is periodic, meaning it repeats itself at regular intervals of time. In the case of a mass on the end of a spring, the spring exerts a force on the mass that acts to bring it back to its equilibrium position. When the mass is at the equilibrium position, the spring force is zero, and the mass is not accelerating. Therefore, the acceleration is maximum at it ends and decreases to zero at equilibrium position.

The instantaneous velocity is zero at the end as it changes direction of motion, though the acceleration is decreasing from end position to equilibrium, velocity is increasing. The instant after equilibrium position, the acceleration is in opposite direction decreasing the velocity till it is zero at the end position.

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Particles q1 = -8.99 uC, q2 = +5.16 uC, and q3 = -89.9 uC are in a line. Particles q1 and q2 are separated by 0.220m and particles q2 and q3 are separated by 0.330m. What is the net force on particle q1?

Answers

Answer: -39200.45 Nm^2/C^2       This is rough work

Explanation:

So, the net force will be felt on particle [tex]\sf{q_1}[/tex] approximately 15.39 N to the right.

Introduction

Hi! Here I will explain the net force of three charges that interacts. Remember that this question can be solved using Coulomb's Law. Coulomb's Law states that the interaction force between two interacted charges will be proportional to the magnitude of the charges that have specific interaction but inversely proportional to the square of the value of the distance between each other. So, with this concept, we can calculate the magnitude of coulomb force with this equation:

[tex] \boxed{\sf{\bold{F = k \cdot \frac{q_1 \cdot q_2}{(r_{12})^2}}}} [/tex]

Formula Used

On this condition, the net force will be felt only by the first charge, so we must define the position of the first charge as a fixed one. So, we can calculate the magnitude of coulomb force concerning the first charge with this equation:

[tex] \boxed{\sf{\bold{\sum F_{q1} = k \cdot (\frac{q_1 \cdot q_2}{(r_{12})^2} + \frac{q_1 \cdot q_3}{(r_{13})^2})}}} [/tex]

With the following condition:

[tex] \sf{\sum F_{q1}}[/tex] = the net force will be felt by the first charge (N)k = constant of coulomb ≈ [tex] \sf{9 \times 10^{9}}[/tex] N.m²/C².[tex] \sf{q_1}[/tex] = first charge (C)[tex] \sf{q_2}[/tex] = second charge (C)[tex] \sf{q_3}[/tex] = third charge (C)[tex] \sf{r_{12}}[/tex] = the distance between first and second charge (m)[tex] \sf{r_{13}}[/tex] = the distance between first and third charge (m)Problem SolvingWe know that:[tex] \sf{q_1}[/tex]= first charge = -8.99 [tex] \sf{\mu C}[/tex] = [tex] \sf{-8.99 \times 10^{-6}}[/tex] C.[tex] \sf{q_2}[/tex]= second charge = +5.16 [tex] \sf{\mu C}[/tex] = [tex] \sf{+5.16 \times 10^{-6}}[/tex] C.[tex] \sf{q_3}[/tex]= third charge = -89.9 [tex] \sf{\mu C}[/tex] = [tex] \sf{-8.99 \times 10^{-5}}[/tex] C.[tex] \sf{r_{12}}[/tex] = the distance between first and second charge = 0.220 m = [tex] \sf{2.2 \times 10^{-1}}[/tex] m.[tex] \sf{r_{13}}[/tex] = the distance between first and third charge = 0.220 + 0.550 m = [tex] \sf{5.5 \times 10^{-1}}[/tex] m.What was asked?[tex] \sf{\sum F_{q1}}[/tex] = the net force will be felt by the first charge = ... NStep by Step:

[See the attached image to know how to solve this problem]

Conclusion:

So, the net force will be felt on particle [tex]\sf{q_1}[/tex] approximately 15.39 N to the right.

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A cannonball is fired with a velocity of 10 m/s upward from horizontal and 30 m/s to the west. What is the angle and direction of the cannonball

Answers

The angle and the direction of the cannonball is calculated to be about 18.435° of west.

Expressed as the ratio between two sides of a right triangle or an angle measure along with a given side, trigonometry can find unknown angle measures and lengths.

Given that,

Velocity of cannonball upward = 10 m/s

Velocity of cannonball to the west = 30 m/s

One of the side lengths reflects the velocity going north (or upward from the horizontal) at 10 m/s. Another side length depicts the speed moving west.

Angle can be calculated by applying trigonometric relation.

tan θ = northward velocity of the cannonball/westward velocity of the cannonball = 10/30

tan⁻¹(tan θ) = tan⁻¹(10/30)

θ = 18.4349°

As, the cannonball was measured from the horizontal, the angle and the direction of the cannonball is about 18.435° of west.

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When building a device or structure, engineers must carefully consider how the materials they choose will change when heated or cooled. List four situations where thermal expansion and contraction could be a problem.

Answers

There are three infrastructure projects wherein thermal expansion & contraction may be a problem are constructing roads, towers, and sidewalks.

What does thermal expansion mean?

The propensity of matter to alter structure, volume, or area in reaction to a temperature change is known as thermal expansion. The average molecular energy of a substance has a monotonic relationship with temperature. Heat makes molecules move more quickly & take up more space, which causes thermal expansion.

What is a good illustration of thermal expansion?

The expansion of heated air is the most obvious example. Heat helps the air expand, making it lighter than the air around it. This causes the hot air to be forced (upward), which causes steam and smoke to rise, warm air balloons to float, and so on.

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Two small objects carry the same amount of charge, and the magnitude of the electric force exerted by one object on the other is 0.10 N when they are held 30 mm apart. If the force magnitude increases when the objects are released and free to move, are the objects of the same charge or opposite charge?

Answers

The objects must be of opposite charge because the electric force of attraction between them increases when they are released and free to move.

This means that the electric force must be a repulsive force that pushes the objects away from each other. Opposite charges attract and as charges repel, the objects must be of opposite charge for the electric force to increase when the objects are released and free to move.

Therefore when two small objects carry the same amount of charge, the magnitude of the electric force exerted by one object on the other is 0.10 N when they are held 30 mm apart. If the force magnitude increases when the objects are released and free to move, then we can say that they are opposite charges.

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If there exists a d such that ad = 0, d > 0, and cd > 0, then is the optimal objective value unbounded?

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With an unbounded optimal solution, the feasible region essentially reaches infinity and even the optimal solution is not constrained by the constraints. Resolution.

This seems to be quite uncommon in real life. One or even more constraints may miss; verify the phrasing of the constraints.

The feasible zone effectively extends into infinity as well as the optimal solution is unbound, meaning the restrictions do not place any upper bounds on it. This is incredibly uncommon in real life. If there are any lacking constraints, kindly verify the formulation of both constraints.

Unbounded functions are those that are neither above nor below bounded by such a finite limit. For instance, because it ranges from to, x is an unbounded function. The unbounded function tax would be defined for any real x with the exception of x(2n+1)2.

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Roller Coaster Physics Gizmo


With no friction, you can use the relationship between potential and kinetic energy to predict the speed of the car at the bottom of this hill from its starting height. To do this start by setting the kinetic and potential energy equations equal to one another. K= U

1/2mv^2 = mhm


a. Use algebra to solver for the speed. V=


b. With the no friction, does the final speed depend on the mass of the car?


c. With no friction, does the final speed depend on the steepness of the hill?


d. What is the final speed of the car if the height of the hill is 55 cm (. 55m)?

Answers

A) Using algebra, the required speed V=(2gh)1/2

B)  With the no friction, the final speed does not  depend on the mass of the car.

C) With no friction, the final speed does depend on the steepness of the hill.

D) The final speed of the car if the height of the hill is 55 cm is 3.569 m/s.

Given that there is no friction considered. Thus, the total energy of the system can be considered as conserved.

Thus, sum of kinetic and potential energy will be constant.

K = U

mu 2   = mgh ......................................(1)

m = mass of object

v = speed of object

g = acceleration due to gravity = 9.8 m/s2

h = height of object with respect to a reference position

Thus for an object at a height from ground, will have its potential energy converted to kinetic energy during it's descent.

(A) from (1)

mv2 = 2mgh

Thus,   v2 = 2gh

            v = (2gh)1/2

this, is the velocity of an object from conservation of energy.

(B) with no friction, the enrgy is taken to be conserved. from the answer of (A), we know,

v = (2gh)1/2.

clearly this equation doesn't depend on mass.

thus, final speed doesn't depend on the mass.

(C) Given height of hill , h = 65cm = 0.65 m

from (1),    v = (2gh)1/2.

the potential energy of car at the hill top = mgh

this potential energy gets entirely converted to kinetic energy at the bottom of hill due to conservation of energy. Thus

v = (2 x 9.8 x 0.65)1/2

   =  3.569 m/s

Thus, the final velocity of the car is 3.569 m/s

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1) compare the isochron map to the topographic map of the seafloor in the previous investigation. which seafloor features are associated with young crust what can you infer from this?
2) how does the age of the seafloor change as you move away from these features?
what can you conclude from this evidence?

Answers

Answer:The isochron map shows the age of the crust of the seafloor, while the topographic map shows the physical features of the seafloor, such as mountains, valleys, and plateaus. Young crust is typically associated with features such as mid-ocean ridges, which are undersea mountain ranges that are formed by the movement of tectonic plates. Mid-ocean ridges are typically associated with high heat flow and volcanic activity, which can result in the formation of new crust.As you move away from these features, the age of the seafloor generally increases. This is because the crust at these features is being continually created through the movement of tectonic plates, while the crust at locations further away from these features is older and has had more time to cool and become more stable.

From this evidence, you can conclude that the seafloor is not a static feature and is constantly being renewed through the movement of tectonic plates and the creation of new crust at mid-ocean ridges. The age of the seafloor can be used to help understand the history and dynamics of the Earth's crust and how it has evolved over time.

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If you are going the same speed, the greater the vehicle _______ the greater the force of impact. florida

Answers

If you are going the same speed, the greater the vehicle, the greater the force of impact on the reacting body and greater the inertia of it and greater is its velocity.

Velocity of any body is the speed by which it is travelling along with the direction it has given. velocity is a vector quantity because it has both magnitude and direction. velocity has both magnitude and direction so it different from speed but its magnitude is same as that of speed and we can calculate both as their magnitude are equal. velocity is the measure of the speed of the moving body by which it is moving in the place from one place to another. velocity and speed are the same but they are different in the essence in the form of their type. speed is scalar and velocity is vector. by this primary information we can consider that primarily that if you are going the same speed, the greater the vehicle, the greater the force of impact on the reacting body and greater the inertia of it and greater is its velocity.

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A skier is pushed from the top of a hill so that he starts moving down the hillside from a height of 100 m with
an initial speed of 0.434 m/s. After traveling 80.4 m, he reaches the bottom of the valley. Due to inertia, he
then continues 70.4 m up another hillside (y = 40 m) and crashes into a snow bank that compressess like a
spring (k = 50 N/m). What is the skiers speed as he crashes into the snow bank? How far does the snow
compress? Assume that you can neglect friction.

Answers

Speed as the skier crashes into the snow bank: 0.434 m/s

Distance the snow compresses: 1.544 m

The speed of the skier as he crashes into the snow bank is the initial speed since there is no acceleration due to the lack of friction. The distance the snow compresses is calculated using the following equation:

compression = (mass x velocity2) / (2 x spring constant)

Therefore, compression = (75 kg x 0.4342) / (2 x 50 N/m) = 1.544 m

A car is traveling for 1638 km for a time period of 6.5 hours. What is the average speed of the car?

Answers

Answer:

252 km/hr

70 m/s

Explanation:

s = d/t

answer in km/hr:

s = 1638 km / 6.5 hr = 252 km/hr

if you need the answer in m/s:

s = ((1638 km)(1000 m/km)) / ( (6.5 hr)(60 min/hr)(60 s/min))

  = 70 m/s

You weigh 600 N.

What would you weigh if the Earth were

seven times as massive as it is and its radius

were three times its present value?

Answer in units of N.

Answers

Your weight would be 85.7N if the Earth were seven times as massive as it is and had a radius three times that of today since the gravitational pull is reduced by seven times with an increase in radius.

Gravitation is a fundamental interaction in physics that produces attraction between all objects with mass or energy. Of the four fundamental interactions, gravity is by far the weakest, being about ten times less than the strong interaction. A body's mass is a fundamental characteristic. Before the discovery of the atom and the field of particle physics, it was generally thought to be tied to the amount of matter in a physical body. x = 600N/7, x = 85.7N

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A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0. 20 km. What is the magnitude of the friction force required to keep the car from sliding?

Answers

The friction force required to keep the car from sliding is 9000N.

Calculation-

mass of car, m = 1000 kg

speed , v = 30 m/s

diameter , d = 0.2 km

r = 100 m

for the car to keep from sliding frictional force is required, ff = m*v^2/r

ff = 1000 * 30^2 /100

ff = 9000 N

the friction force required is 9000 N

What kind of force does an automobile exert on a straightaway?

Simply put, an unbanked curve (or turn) is a curve that is flat on the ground (parallel to the horizontal). A force of friction applies to the car as it drives along one of these curves, keeping it rotating in a circular course.

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Why does torque increase with distance?

Answers

Torque increases with distance because torque is defined as the force acting on an object multiplied by the distance from the axis of rotation. The farther the force is applied from the axis of rotation, the greater the torque will be.

To understand this, imagine a lever with a weight on the end of it. If the weight is placed closer to the pivot point of the lever, it will take less force to lift the weight than if the weight is placed farther away from the pivot point. This is because the farther the weight is from the pivot point, the greater the torque will be. The same is true for any object that is being rotated around an axis - the farther the force is applied from the axis of rotation, the greater the torque will be.

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Suppose a 51-N sled is resting on packed snow. The coefficient of kinetic friction is 0.13. If a person weighing 680 N sits on the sled, what force is needed to pull the sled across the snow at constant speed

Answers

95 N much force is needed to pull the sledge at a constant speed.

Explain force.

A force in physics is an effect that has the power to alter what an object moves. Every object with mass might travel faster or at a different rate as a result of a force. A push or a pull made natural sense when describing force. Since forces are vector quantities, they own both size and direction.

The sled is being subjected to the following horizontal forces:

F = applied force.

Fr = friction force.

And the following vertical forces:

N = normal force.

W =combined weight of the person and the sled

The second law of Newton states:

F = m * a

where the object's mass is "m" and its acceleration is "a"

So, in the horizontal direction:

F - Fr = m *a

Determining the force, F, required to keep the sled moving at a constant pace (acceleration = 0). Then,

F - Fr = 0

F = Fr

The applied force must have a magnitude equal to or greater than the friction force.

The friction force is calculated as follows:

Fr = μ · N

Where μ is coefficient and N represent the normal force, respectively.

Let's use Newton's second law with in vertical direction to determine the normal force:

F = N - W

   = m · a

Keep in mind that the sledge is not accelerated vertically so that a = 0:

N - W = 0

N = W

The mass is the total of a weight of a person plus the mass of the sled, and the force is equivalent to the weight:

W = 51 N + 680 N = 731 N

Then:

N = 731 N

Fr = 0.13 · 731 N

Fr =  N

Then

F = 95.03 N

As a result, 95 N of force is necessary to pull a sled at a steady pace.

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1. A 10 kg rubber block sliding on a concrete floor (μ=0. 65) 2. A 248 kg object moving at 19 m/s comes to stop over a distance of 38 m. What is the coefficient of kinetic friction between the surfaces? HURRY 50pts and dont take the points just answer the question

Answers

The coefficient of kinetic friction between the surfaces is 0.48.

What is kinetic friction?

Friction is simply defined as the force that prevents a sliding object from moving. Kinetic friction is a natural phenomenon that disrupts the motion of two or more objects. The force acts in the opposite direction of the motion of the object.

When we stop a car, we use brakes, which is where friction comes into play. When walking, friction is to thank once more. But when we have to stop in the middle of a puddle, things become more difficult because friction is less and cannot help as much.

Solve for a using K3:    [tex]$v^2=v_0^2+2 a d$\\[/tex]

[tex]& 19^2=0_0^2+2 a 38 \\& a_x=-4.75 \mathrm{~m} / \mathrm{s}^2[/tex]

[tex]\begin{aligned}& \text F_{\text {net }}=-f_k=m a_x, \text { and } \\& -f_k=\mu_k F_n=\mu_k m g, \text { so we can say } \\& -m a_x=\mu_k \cdot m g \\& \mu_k=\frac{m a_x}{m g}=\frac{-a_x}{g} \\& \mu_k=\frac{-(-4.75)^m / s^2}{9.8}=0.48\end{aligned}$$[/tex]

Thus, The coefficient of kinetic friction between the surfaces is 0.48.

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To report the position of an object, first choose a reference point. Then, describe the of the object with respect to the reference point. Next, measure the from the reference point to the object.

Answers

We will understand the given concept through an example

The rat serves as your point of reference if you mentioned that the rabbit was to the left of it. You are expressing its position when you specify the distance and direction. The present separation and direction of an object from a reference point is its position.

When something moves with relation to a reference frame, like when a passenger gets off an airline or a lecturer gets up to leave, the object's position changes.the right in relation to a whiteboard. Displacement describes this shift in location.Every location helps you to achieve various location positions.Your position is defined by how you describe your place. The separation in the first illustration is one meter. The reference point helps to locate each and every point.

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the effect of a stress raiser is more significant for which of the following types of materials?
a Ductile
b.Gaseous
c Volatile
d. Brittle

Answers

Option A is the correct Answer. The effect of a stress raiser is more significant for Ductile materials.

Ductility is the ability of a material to be drawn or plastically deformed without fracture. The ductility of steels varies depending on the types and levels of alloying elements present.

Examples: aluminum, copper, and magnesium alloys.

Uses of  Ductile materials:

Ductility is mostly used in metalworking, as materials that crack, break or shatter under stress cannot be manipulated using metal-forming processes such as hammering, rolling, drawing, or extruding.High degrees of ductility occur due to metallic bonds, which are found predominantly in metals; this leads to the common perception that metals are ductile in general.The ductility of steel varies depending on the alloying constituents. Increasing the levels of carbon decreases ductility.

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How are static and passive stretching similar?

Answers

Static stretching and passive stretching are very similar in that they both involve moving the limbs into and holding the stretch posture.

Static stretching involves the individual holding their own limbs in place, while passive stretching involves a partner assisting with the movement and holding of limbs. When you stretch to your limit and hold that position, you are engaging in a form of stretching known as static stretching. Passive stretching, on the other hand, is also a form of static stretching; however, during this technique, you remain relaxed and do not actively participate in increasing your range of motion.

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Select True or False: The mass of a nucleus is always less than the sum of the masses of the nucleons.

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It is untrue to say that a nucleus's mass is always smaller than the total mass of its nucleons.

Why is nucleus mass lower than nucleon mass?

The nucleons near the edge of their masses give binding energy to maintain the nucleons inside the nucleus.

Why is a nucleus' mass always smaller than the total?

The protons and neutrons move in closer together to a distance of 10–14 m during the creation of a nucleus. The mass of the nucleons is sacrificed in order to spend the energy needed for the task.

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What is the work applied to a 1000kg car when it is pushed 1000m up along a 10- degree incline at a constant velocity. Assume the coefficient of friction between the car and the ramp is 0.30.

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The required work applied to a 1000 kg car when it is pushed 1000 m up along a 10- degree incline at a constant velocity is calculated to be 4.6 × 10⁶ J.

Given that,

Mass m = 1000 kg

Displacement d = Δx = 1000 m

Angle of inclination θ = 10°

Coefficient of friction μ = 0.3

The equation for work done is known to be,

W = F × d

We must divide the force of gravity into its component parts and examine the side acting in the x-direction in order to calculate the force of gravity in that direction.

F gx direction = F g sinθ = mg sinθ

The force of friction is directly related to  μ times the normal force.  The normal force in this situation is equivalent to the y component of the force of gravity.

F gy direction = F g cosθ = mg cosθ

We know force of friction, Ff = μ FN

FN = F gy direction

Ff = μ mg cosθ

We know that our force applied must balance both of these forces,

F applied = mg sinθ + μ mg cosθ

As, W = F × d = (mg sinθ + μ mg cosθ) d = ( 1000× 9.8 sin10° + 0.3 × 1000× 9.8 cos10°) 1000 = (9800 sin10° + 2940 cos10°) 1000 = (0.17 × 9800 + 0.985 × 2940) 1000 = (1666 + 2895.9) 1000 = 4561.9 × 10³ J = 4.6 × 10⁶ J

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What are two way in which early psychological research was biased? What’s an example of bias in psychology?

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Psychological research gives us information about the research that psychologists conduct for systematic study and for analysis of the experiences and behaviors of individuals or groups.

Give the two methods on which early psychological research was based.

There are two ways in which psychological research is based and that is the research being androcentric and ethnocentric. Examples of this include the theory of personality and the attachment of the theory.

There are many more biases is well in psychological research which are Citation bias, Language bias, and Outcome reporting bias. The citation bias includes whether the research receives citations or not. Language bias includes when research is published in one language over another. The outcome reporting bias includes that when only some outcomes are reported from a study.

So we can conclude that there are two ways in which early psychological research was biased and that are; when research is androcentric and ethnocentric.

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2. A 1.0 kg roller coaster is at the top of a 25m hill, and has a velocity of 8 m/s.

a. Determine the velocity of the roller coaster when it is halfway down the hill, at height of 12.5 m. (Ignore Friction)

b. What is the velocity of the roller coaster at the bottom of the hill assuming a force due to friction of 3.0 N, and the length of the hill is 45.0 m.

SHOW ALL WORK!

Answers

The car's acceleration can be calculated as: a = v2/R = (18.0 m/s) 2/(12.0 m) = 27.0 m/s2. The conventional method can be used to determine net force: 13500 N is equal to Fnet = m•a = (500 kg)•(27.0 m/s2).

What happens to a roller coaster's speed as it climbs a hill?

It may slow down or decelerate when traveling uphill or straight ahead. Roller coasters accelerate as they travel down hills due to the force of gravity dragging them downward.

How do roller coasters use math?

Calculus and other fundamental mathematical concepts are used to calculate the maximum speed, the angles of ascent and descent, and the height necessary for the automobile to climb the next hill. The roller coaster's safety is further ensured by these calculations.

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