A 1300 kg car is driving east on 290 at 65 mph. What is the car's momentum?

The KEf is less than KEi, we can conclude that this is not an elastic collision. Some kinetic energy was lost during the collision, possibly due to friction between the pucks or deformation of the pucks. The final velocity of the second puck is also 0.382 m/s, since the two pucks stick together after the collision.

What is the conservation of kinetic energy?

The conservation of kinetic energy is a fundamental principle in physics that states that in an isolated system, the total kinetic energy of the system remains constant, provided that no external forces act on the system. This means that the kinetic energy of the system is conserved during any process or interaction between the objects in the system. In other words, the total kinetic energy of the system before and after the interaction is equal.

We can use momentum conservation and kinetic energy conservation to solve this problem.

First, let's find the initial momentum of the system. Since the second puck is stationary, its momentum is zero. The momentum of the first puck is:

p1i = m1v1i = (0.170 kg)(1.50 m/s) = 0.255 kg m/s

The system's initial total momentum is:

pTotali = p1i + p2i = 0.255 kg m/s

After the collision, the first puck is moving in a direction that is 30 degrees away from the blue line at a speed of 0.75 m/s. Let's break this velocity into its x and y components:

vx1f = v1f cos(30) = 0.75 m/s cos(30) = 0.65 m/s

vy1f = v1f sin(30) = 0.75 m/s sin(30) = 0.375 m/s

Now, let's use conservation of momentum to find the final velocity of the second puck. Since the collision is inelastic, the two pucks will stick together after the collision, so their final velocities will be the same.

pTotalf = pTotali

(m1 + m2)vf = m1v1f + m2v2f

Replacing the values we know, we get:

(0.34 kg)vf = (0.170 kg)(0.65 m/s) + (0.170 kg)(0 m/s)

Solving for vf, we get:

vf = 0.382 m/s

The final velocity of the second puck is also 0.382 m/s, since the two pucks stick together after the collision.

To check whether this is an elastic collision, we can use the conservation of kinetic energy. The system's initial kinetic energy is:

(1/2)m1v1i2 + (1/2)m2(0 m/s)2 = 0.0574 J

The system's total kinetic energy is:

KEf = (1/2)(m1 + m2)vf² = 0.0277 J

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The following is example of uniform acceleration, that is : motion under force.

Uniform acceleration refers to constant rate of change in velocity, which means that acceleration of an object is constant over time. Motion under force can result in uniform acceleration if the force is constant and object is free to move in straight line.

Motion under gravity and motion under power can both result in acceleration but they are not necessarily uniform. For example,  object falling under gravity experiences constantly increasing acceleration due to increasing speed and gravitational force acting on it, which means acceleration is not uniform.

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We can see here that the trend of the data shows in each trial, the number of bread and cheese used drastically reduced or are used up after each reaction.

The data represents  the Law of Conservation of Mass by showing that if you add two elements you will have one product.

The Law of Conservation of Mass is a fundamental principle in chemistry, which states that the total mass of the reactants in a chemical reaction is equal to the total mass of the products.

In other words, the total mass of matter in a closed system remains constant during a chemical reaction, and no matter is created or destroyed.

The part that completes the question is seen below:

9. Conduct the SAME trials as you did above, but this time count the number of ingredients you started with and what you ended up with. You are NOT counting sandwiches, but all the individual ingredients.

Trial Number                                  1          2         3       4        5    

#of Bread Before Reaction           5         6         3      5         6

#of Cheese Before Reaction        3         4         4      2         5

#of Bread After Reaction              1          0         1       1          0

#of Cheese After Reaction           1          1          3      0         2

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1. Conjecture: In triangle ABC and A'B'C', all corresponding segments are parallel if the scale factor is greater than or equal to 1.

Let ABC and A'B'C' be two triangles with corresponding sides AB and A'B', BC and B'C', and AC and A'C'. Assume that the scale factor between the two triangles is greater than or equal to 1.

To prove that all corresponding segments are parallel, we need to show that the slopes of these segments are equal.

Consider segment AB and A'B'. Let (x₁, y₁) and (x₂, y₂) be the coordinates of points A and B, and (x₁', y₁') and (x₂', y₂') be the coordinates of points A' and B', respectively. Then, the slope of segment AB is (y₂ - y₁)/(x₂ - x₁), and the slope of segment A'B' is (y₂' - y₁')/(x₂' - x₁').

Since the scale factor is greater than or equal to 1, we have |AB| <= |A'B'|, |BC| <= |B'C'|, and |AC| <= |A'C'|.

Without loss of generality, assume that |AB| <= |A'B'|. Then, we have:

|x₂ - x₁| <= |x₂' - x₁'|, and |y₂ - y₁| <= |y₂' - y₁'|.

Thus, we can conclude that (y₂ - y₁)/(x₂ - x₁) = (y₂' - y₁')/(x₂' - x₁'), which implies that segment AB is parallel to segment A'B'.

Similarly, we can prove that segment BC is parallel to segment B'C' and segment AC is parallel to segment A'C'.

Therefore, we have shown that all corresponding segments are parallel when the scale factor is greater than or equal to 1.

3. If the scale factor is less than one, the proof would need to be modified. In this case, we would have |AB| >= |A'B'|, |BC| >= |B'C'|, and |AC| >= |A'C'|. As a result, we would need to adjust the inequality signs in the proof accordingly. Specifically, we would need to reverse the signs of the inequalities involving the absolute values of the differences between the x and y coordinates of the points in each pair of corresponding segments.

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Answer:25 c

Explanation:

(a) Frequency of the tuning fork is 686 Hz.

(b) Minimum value to hear a resonance is 0.25 m.

(c) The next level above 0.625 to hear a resonance is 0.875 m.

Resonance frequency is defined as the natural frequency at which the medium vibrates with maximum amplitude.

Here,

(a) Velocity of sound, v = 343 m/s

    Distance, L₁ = 0.375 m

                L₂ = 0.625 m

     From the equation,

    ΔL = 1/2 λ

    λ = 2ΔL = 2(0.625 - 0.375)

    λ = 0.5 m

  Therefore,

  frequency, f = v/λ

  f = 343/0.5

  f = 686 Hz

(b) Minimum value to hear a resonance = λ/2 = 0.25 m

(c) Next level above 0.625 to hear resonance,

L₃ = 0.625 + λ/2 = 0.625 +(0.5/2)

L₃ = 0.875 m

Hence,

(a) Frequency of the tuning fork is 686 Hz.

(b) Minimum value to hear a resonance is 0.25 m.

(c) The next level above 0.625 to hear a resonance is 0.875 m.

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Your question was incomplete. Attaching the image here.

If the stone rises to a height of 5 m above the point of projection, then the value of u is 49 m/s, the time the stone takes from projection until it reaches ground level is 5 seconds, and the speed of the stone at ground level is 98 m/s.

To find the value of u,

v = u + at (velocity (v) ,the initial velocity (u), acceleration (a), and time (t))

At the highest point, the velocity of the stone becomes zero,

0 = u - gt

t = u/g

this value of t is put into the equation for displacement,

5 m = 120 m + ut - [tex]0.5gt^2[/tex]

Substituting t = u/g,

5 m = 120 m + u(u/g) - [tex]0.5g(u/g)^2[/tex]

u = √(2gh) = √(2 x 9.8 [tex]m/s^2[/tex] x 115 m) = 49 m/s

b) v = u + at

-u = 49 m/s - gt

t = (49 m/s)/g + √(([tex]49 m/s)^2[/tex] + 4gh)/g

t = 5 s

c)

v = u + at

-u = 49 m/s - g(5 s)

u = 49 m/s + 9.8[tex]m/s^2[/tex] x 5 s

u = 98 m/s

Hence, the value of u is 49 m/s, the time the stone takes from projection until it reaches ground level is 5 seconds, and the speed of the stone at ground level is 98 m/s.

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When the rest of a half-filled container containing a steel ball floating on the surface of the mercury is filled with water, the ball remains partially submerged in mercury but not as deeply as before.

Whether a substance will float or sink in a liquid is determined by its density relative to the density of the liquid.

If the density of the substance is less than the density of the liquid, it will float because it is less dense than the liquid and will experience a buoyant force that is greater than its own weight. If the density of the substance is greater than the density of the liquid, it will sink because it is denser than the liquid and will experience a buoyant force that is less than its own weight.

For example, water will float on mercury since it is less dense than mercury.

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Complete question:

Consider a steel ball floating on the surface of the mercury in a half-filled container.

What happens when the rest of the container is filled with water? Mercury is denser than steel, and both are denser than water. Mercury and water do not mix.

1. The ball remains partially submerged in mercury to exactly the same depth as before.

2. The ball Boats to the surface of the water.

3. Mercury floats on top of the water, and the ball floats on the mercury's surface

4. The ball remains partially submerged in mercury but not as deeply as before.

5. The ball sinks to the bottom of the container

6. The ball remains partially submerged in mercury to a greater depth than before.

The key difference between the two types of generators is that DC generators produce a constant and unidirectional output while AC generators produce a varying and alternating output.

The output of a DC generator produces a steady, unidirectional current that flows in one direction. The plot of its output voltage versus time would be a straight line with no fluctuations or changes in direction.

On the other hand, an AC generator produces an alternating current that changes direction periodically, usually at a fixed frequency. The plot of its output voltage versus time would be a sinusoidal waveform with a repeating pattern of positive and negative values.

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Here, the total momentum before and after the release of the projectile is same. From this concept the final velocity of the cannon after the release will be 0.00433 m/s.

Momentum of an object is the product of its mass and velocity. During a collision, the momentum after and before the collision is constant. This can be applied in the case of a projectile from the cannon.

Here, the initial momentum of the cannon and the ball is zero, because they are at rest.

After, projectile the total momentum momentum is zero itself according to conservation of momentum.

momentum of the ball = 0.05 kg × 0.1 m/s cos 30 + 1 kg × v = 0

Here v is the velocity of the cannon.

then v = 0.05 kg × 0.1 m/s cos 30  / 1 kg

          = 0.0043 m/s.

Therefore, the velocity of the cannon after the projectile is released is 0.0043 m/s.

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Answer:

1. To solve this problem using conservation of momentum, we can use the equation:

(m1 * v1) + (m2 * v2) = (m1 + m2) * vf

where m1 and v1 are the mass and velocity of the first object, m2 and v2 are the mass and velocity of the second object, and vf is the final velocity of the two objects after they stick together.

Plugging in the given values, we get:

(2 kg * 5 m/s) + (3 kg * 0 m/s) = (2 kg + 3 kg) * vf

Simplifying the equation, we get:

10 kg m/s = 5 kg * vf

vf = 2 m/s

Therefore, the final velocity of the two objects after the collision is 2 m/s.

2. Using the same equation as above and plugging in the given values, we get:

(1000 kg * 10 m/s) + (500 kg * 0 m/s) = (1000 kg + 500 kg) * vf

Simplifying the equation, we get:

10,000 kg m/s = 1500 kg * vf

vf = 6.67 m/s

Therefore, the final velocity of the two cars after the collision is 6.67 m/s.

3. Before the boy jumps off the bike, the total momentum of the system is:

(50 kg + 10 kg) * 5 m/s = 300 kg m/s

After the boy jumps off, the total momentum of the system is:

50 kg * v_bike

Using conservation of momentum, we can equate the two and solve for v_bike:

(50 kg + 10 kg) * 5 m/s = 50 kg * v_bike

Simplifying the equation, we get:

300 kg m/s = 50 kg * v_bike

v_bike = 6 m/s

Therefore, the velocity of the bike after the boy jumps off is 6 m/s.

4. Using the same equation as in problem 1 and plugging in the given values, we get:

(75 kg * 8 m/s) + (85 kg * -8 m/s) = (75 kg + 85 kg) * vf

Simplifying the equation, we get:

0 = 1600 kg * vf

vf = 0 m/s

Therefore, the final velocity of the two players after the collision is 0 m/s.

5. Using conservation of momentum and plugging in the given values, we can set up the equation:

(2 kg * 5 m/s) + (1 kg * 0 m/s) = (2 kg + 1 kg) * 2 m/s + (1 kg * vf)

Simplifying the equation, we get:

10 kg m/s = 5 kg * 2 m/s + vf

vf = 0 m/s

Therefore, the final velocity of the 1 kg ball after the collision is 0 m/s.

6. To solve this problem, we can use the conservation of momentum equation again:

(500 kg * 500 m/s) = (450 kg * vf_r) + (50 kg * 1000 m/s)

where vf_r is the final velocity of the rocket after releasing the satellite.

Simplifying the equation, we get:

250,000 kg m/s = 450 kg * vf_r

vf_r = 555.56 m/s

Therefore, the velocity of the rocket after releasing the satellite is 555.56 m/s.

7. Using the same conservation of momentum equation and plugging in the given values, we get:

(5 kg * 10 m/s) + (2 kg * 0 m/s) = (5 kg + 2 kg) * vf

Simplifying the equation, we get:

50 kg m/s = 7 kg * vf

vf = 7.14 m/s

Therefore, the final velocity of the 5 kg mass after the collision is 7.14 m/s.

8. Using the same conservation of momentum equation and plugging in the given values, we get:

(1 kg * 2 m/s) + (2 kg * 0 m/s) = (1 kg + 2 kg) * 1 m/s + (2 kg * vf)

Simplifying the equation, we get:

2 kg m/s = 3 kg + (2 kg * vf)

vf = -0.5 m/s

Therefore, the final velocity of the 2 kg mass after the collision is -0.5 m/s. This negative velocity indicates that the mass is moving in the opposite direction to the initial direction.

According to the solving the final velocity of the two objects after they stick together is 1.67 m/s.

The velocity of a body determines whether it is moving away from the ground or towards an object. Speed is often a scalar quantity. Velocity is a vector quantity in its most basic form. It gauges the rate of change in a distance. It relates to how quickly displacement is altering.

the objects stick together after the collision, we can use conservation of momentum to solve for the final velocity of the combined mass:

m1 * v1 + m2 * v2 = (m1 + m2) * vf

where m1 and v1 are the mass and velocity of the first object, m2 and v2 are the mass and velocity of the second object, and vf is the final velocity of the combined mass.

Substituting the given values:

(2 kg) * (5 m/s) + (3 kg) * (0 m/s) = (2 kg + 3 kg) * vf

Simplifying and solving for vf:

vf = 1.67 m/s

the final velocity of the two objects after they stick together is 1.67 m/s.

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The letter that represents the various descriptions is as follows:

5.) Angle of incidence: B

6.) Object: D

7.) Plain mirror:A

8.) Reflection: E

9.) Angle of reflection: C

10.) Normal: F

The law of reflection in a plain mirror states that the angle of reflection is equal to the angle of incidence.

The diagram explains the law of reflection on a plain mirror because it shows the reflection of the object D as E after creating an angle of incidence B on the plane mirror surface A which is equal to the angle of reflection C.

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Answer:

1.3 m/s

Explanation:

To solve this problem, we can use the equations of motion for a freely falling body with constant acceleration due to gravity (g = 9.81 m/s²).

a. The time for which it is above a height of 63m:

We can use the following equation of motion to find the time t it takes for the particle to reach a height of 63 m above the ground:

y = uyt + (1/2)gt²

where y is the displacement (height) of the particle, uy is the initial velocity in the y-direction (upward), and t is the time.

At the highest point of its trajectory, the particle will have zero velocity, so we can use the initial velocity as uy = 36 m/s. We want to find the time t when the particle reaches a height of y = 63 m, so we can rearrange the equation to solve for t:

t = (-uy ± sqrt(uy² + 2gy))/g

where the ± sign indicates that there are two possible solutions, one for when the particle is going up (positive root) and one for when the particle is coming down (negative root).

Plugging in the values for uy, g, and y, we get:

t = (-36 ± sqrt(36² + 2*(-9.81)*63))/(-9.81)

t = 6.53 s (upward)

b. The speed which it has at this height on its way down:

When the particle reaches a height of 63 m on its way down, we can use the following equation to find its velocity vy:

vy² = uy²+ 2g(y - yo)

where yo is the initial height (0 m) and uy is the initial velocity in the y-direction (upward).

We can plug in the values for uy, g, and y, and solve for vy:

vy²= 36²+ 2(-9.81)(63 - 0)

vy = 32.2 m/s (downward)

c. The total time of flight:

The total time of flight is twice the time it takes for the particle to reach the maximum height, since it takes the same amount of time to come back down to the ground. Therefore, the total time of flight is:

t_total = 2t_upward = 2(6.53) = 13.06 s

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The mechanical advantage of the lever will be 4.

The distance the gardener pushed the shovel will be 0.8 m.

a) The mechanical advantage of the lever can be calculated using the formula:

MA = output force/input force

In this case, the output force is the weight of the rock (200 N) and the input force is the force applied to the shovel (50 N). Therefore:

MA = 200 N / 50 N = 4

So the mechanical advantage of the lever is 4.

b) The distance that the gardener pushes the shovel can be calculated using the formula:

output distance / input distance = input force / output force

In this case, the output distance is the distance that the rock is lifted (0.20 m) and the input distance is the distance that the gardener pushes the shovel (unknown). We know the input force (50 N) and the output force (200 N), so we can set up the equation:

0.20 m / x = 50 N / 200 N

Simplifying and solving for x:

0.20 m * 200 N / 50 N = x

0.8 m = x

Therefore, the gardener pushes the shovel down 0.8 m.

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The total initial mechanical energy is 590,650 J.

Work done on the projectile is 155,436 J.

Speed of the projectile is 115.4 m/s.

(a) The initial total mechanical energy of the system of the projectile and the earth can be found using the equation:

E = KE + PE

where E is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy. Initially, the projectile is at a height of 116 m, so the potential energy is:

PE = mgh

PE = (54.0 kg)(9.81 m/s²)(116 m)

PE = 61,450 J

The initial speed of the projectile is 1.40 x 10² m/s, so the initial kinetic energy is:

KE = (1/2)mv²

KE = (1/2)(54.0 kg)(1.40 x 10² m/s)²

KE = 529,200 J

Therefore, the initial total mechanical energy is:

E = KE + PE

E = 529,200 J + 61,450 J

E = 590,650 J

The initial total mechanical energy of the system of the projectile and the earth is 63,014 J.

(b) At the maximum height, the potential energy of the projectile is:

PE = mgy

PE = (54.0 kg)(9.81 m/s²)(320 m)

PE = 169,517 J

At the maximum height, the kinetic energy of the projectile is:

KE = (1/2)mv²

KE = (1/2)(54.0 kg)(99.2 m/s)²

KE = 265,697 J

The total mechanical energy of the projectile at the maximum height is:

E = KE + PE

E = 169,517 J + 265,697 J

E = 435,214 J

The work done by air resistance is the difference between the initial total mechanical energy and the total mechanical energy at the maximum height:

W = Ei - Ef

W = 590,650 J - 435,214 J

W = 155,436 J

The amount of work done on the projectile by air resistance is 155,436 J.

(c) Let Wup be the work done by air resistance when the projectile is going up, and let Wdown be the work done by air resistance when the projectile is going down. According to the problem, Wdown = (3/2)Wup. Use the principle of conservation of energy to relate the speed of the projectile at its maximum height to its speed just before hitting the ground:

Ef = Ei

KEf + PEf = KEi + PEi

At the maximum height, the kinetic energy of the projectile is zero, so:

PEf = Ei - PEi

PEf = 590,650 J - (54.0 kg)(9.81 m/s²)(320 m)

PEf = 421,133 J

The total mechanical energy just before hitting the ground is the same as the total mechanical energy at the maximum height, so:

KEf + PEf = KEi + PEi

0 + 421,133 J = (1/2)mv² + 61,450 J

(1/2)(54.0 kg)v² = 359,683 J

v² = 13,321.59

v = 115.4 m/s

Therefore, the speed of the projectile immediately before it hits the ground is 115.4 m/s.

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Water exists in three different states, namely solid ice, liquid and gaseous vapour. There are two different formulae, depending upon whether there is a loss or gain of heat, or change of state.

The specific heat capacity is the amount of heat absorbed per unit mass of the substance for a temperature increase of one unit. Its units are given by J/kg K or cal/kg °C. It is useful in calculating the amount of heat lost or gained by a body (Q). The formula for Q is:

Q = m*c*ΔT

where, m is the mass

c is the specific heat capacity and,

ΔT is the change in the temperature of the body

This formula is used to measure the amount of energy required to raise or decrease the temperature of water without a change in its state.

On the other hand, latent heat is the energy released or absorbed when there is a change of state of matter. It is denoted by L. For a change of state, the heat energy Q is given by:

Q = m*L

Note that there is no change in temperature during change of state of matter, and only heat is released or absorbed.

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Newton’s second law says that when an unbalanced force is applied to a mass, it causes it to accelerate.

What is Newton's law?

Newton's laws are a set of three fundamental principles that describe the behavior of objects in motion, developed by the physicist Sir Isaac Newton in the late 17th century. The laws are as follows:

The first law, also known as the law of inertia, states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity, unless acted upon by an external force.

The second law, also known as the law of acceleration, states that the acceleration of an object is directly proportional to the net force applied to it, and inversely proportional to its mass. This can be expressed mathematically as F = ma, where F is the net force applied, m is the mass of the object, and a is the resulting acceleration.

The third law, also known as the law of action-reaction, states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.

Together, these laws provide a framework for understanding the behavior of objects in motion, and have been applied in countless areas of science and engineering.

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If I mix hot and cold water together, then the final temperature of the water will be somewhere in between the initial temperatures of the hot and cold water.

Temperature is a measure of heat energy, which is defined as the average kinetic energy of particles in a substance. It is measured using a thermometer, and is expressed in either Celsius (°C) or Fahrenheit (°F). Temperature is an important physical property as it affects the rate of chemical reactions, the speed of sound and light, and the behavior of matter.

It is also used to measure the intensity of heat energy in terms of the amount of thermal energy generated or absorbed by an object or material. Temperature is a fundamental physical property and is used to quantify the energy of a system. Temperature increases when energy is added to a system, and decreases when energy is removed.

If I mix hot and cold water together, then the final temperature of the water will be somewhere in between the initial temperatures of the hot and cold water.

Independent Variable: Temperature of hot and cold water.

Dependent Variable: Final temperature of the water.

Controlled Variables:

Amount of hot and cold water, type of cups, thermometer, and measuring cups.

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1) There are many handwriting characteristics that are used to identify forgeries, including line quality, pen pressure, slope, letter size and shape, and more.

Pressure is a physical force that is exerted on an object or area by another object or area. It is measured in terms of force per unit area and can be calculated using the formula: pressure = force/area. Pressure is an important factor in many areas of science, engineering, and everyday life.

2) Checks have embedded features such as watermarks, microprinting, and holograms to prevent forgery.

3) True. A person's handwriting is compared to several exemplars to determine if the handwriting is genuine or a forgery.

4) True. Some forgers use chemicals to age paper to make it appear older than it is.

5) The U.S. Department of the Treasury is charged with the security of U.S. currency.

6) False. People often spend time forging documents from famous people in order to make money.

7) U.S. currency has many security features, including watermarks, color-shifting ink, invisible fibers, and microprinting.

8) The main difference between check forgery and literary forgery is that check forgery is primarily concerned with creating or altering a financial document, while literary forgery is primarily concerned with creating or altering a work of literature.

9) Criminals make forged literary pieces look older by using aging techniques such as adding age spots

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Angular velocity of the drum at this speed having 15m diameter is 3.6 rad/s

Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force.

frictional force is directly proportional to the Normal(N).

i.e. [tex]F_{fri}[/tex]∝ N

[tex]F_{fri}[/tex] = μN

where μ is called as coefficient of the friction. It is a dimensionless quantity.

When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.

Given,

Diameter of the drum D = 10m , Radius r = 5m

Coefficient of static friction μ = 0.15

To stay everyone pinned against the wall of drum. Frictional Force must be equal to weight mg which are opposite to each other.

μN = mg ........1)

Centrifugal acceleration = Normal

mv²÷r = N

With this equation 1 becomes

μmv²÷r = mg

v² = rg÷μ

v² = 5m*9.8m/s ÷ 0.15

v² = 326.6

v=  = 18.07 ~ 18 m/s

Hence minimum linear velocity required for the drum  is 18m/s.

for angular velocity of the drum, V=rω

ω = v÷r

ω = 18m/s ÷ 5m

ω = 3.6 rad/s

Hence angular velocity of the drum at 18m/s speed is 3.6 rad/s

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Electromagnetic and electrical control equipment is used in a wide range of industrial and commercial applications to control the operation of electric motors, lighting, and other electrical loads.

Control relays, contactors, and motor starters are key components of electromagnetic and electrical control equipment.

They are used to control the operation of electric motors, lighting, and other electrical loads.

Solid-state devices have become increasingly popular in recent years due to their compact size, faster response times, and lower power consumption. NEMA and IEC establish standards for electrical equipment, and electrical components are often rated according to these standards.

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Electromagnetic induction is the process of generating an electric current by moving a conductor through a magnetic field.

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Answer:

0.0312J

Explanation:

Given:

Want: Elastic potential energy stored in the spring

Solve:

The elastic potential energy stored in the spring can be calculated using the following formula:

Elastic potential energy = 0.5 * k * (L - L0)^2

Substituting the given values into the equation:

Elastic potential energy = 0.5 * 51.0 N/m * (0.150 m - 0.115 m)^2

Elastic potential energy = 0.5 * 51.0 N/m * (0.035 m)^2

Elastic potential energy = 0.5 * 51.0 N/m * 0.001225 m^2

Elastic potential energy = 0.0312 J

Therefore, the elastic potential energy stored in the spring when its length is 0.150 m is 0.0312 J.

Answer:

I=4.5A,t=25minutes Q=?

We know that I=Q/T

Q=25×4.15=

112.5c

Explanation:

The values for the work done to the nearest 10 is as follows:

(a) The work done by the applied force can be calculated using the formula:

WA = FAdcosθ,

where FA is the applied force, d is the distance moved, θ is the angle between the force and the displacement, and cosθ is the cosine of that angle.

In this case, FA = 216 N, d = 71.0 m, and θ = 30.0° below the horizontal. Therefore,

WA = (216 N)(71.0 m)cos(-30.0°)

WA = 13,200 J.

(b) The force of gravity on the box is given by Fg = mg

The work done by the force of gravity is given by Wg = Fgd,

Fg = (51.0 kg)(9.81 m/s^2)

Fg = 500.31 N, and the box does not move vertically, so d = 0.

Therefore, the work done by the force of gravity is zero.

Wg = 0 J.

(c) The normal force on the box is equal in magnitude and opposite in direction to the force of gravity, so FN = Fg. Since the box does not move vertically, the normal force does not do any work.

Therefore,

WN = 0 J.

(d) The force of friction on the box can be calculated using the formula:

FF = μFN,

where μ is the coefficient of kinetic friction between the floor and the box, and FN is the normal force on the box.

In this case, μ = 0.100 and FN = Fg = 500.31 N. Therefore,

FF = (0.100)(500.31 N) = 50.03 N

The work done by the force of friction can be calculated using the formula:

WF = FFd

In this case, the box moves a distance of 71.0 m.

Therefore,

WF = (50.03 N)(71.0 m)

WF = 3,550 J.

(e) The net work on the box is the sum of the works done by each individual force.

Therefore,

WNet = WA + Wg + WN + WF = 13,200 J + 0 J + 0 J + 3,550 J

WNet = 16,750 J.

(f) Alternatively, we can find the net work by first finding the net force on the box.

The horizontal component of the applied force is FAcosθ, where θ = 30.0° below the horizontal.

Therefore,

FNet,x = FAcosθ - FF = (216 N)cos(30.0°) - (0.100)(500.31 N)

FNet,x = 91.10 N

The vertical component of the applied force is FAsinθ, where θ = 30.0° below the horizontal. Therefore,

FNet,y = Fg - FAsinθ =  (51.0 kg)(9.81 m/s^2) - (216 N)sin(30.0°)

FNet,y = 247.16 N

The net force on the box can be found using the Pythagorean theorem:

FNet = √(FNet,x² + FNet,y²)

FNet = √((91.10 N)² + (247.16 N)²)

FNet = 264.36 N

The direction of the net force is given by the angle θ between FNet,x and FNet:

θ = atan(FNet,y / FNet,x)

θ = atan(247.16 N / 91.10 N

θ = 70.9°

The net work done on the box is given by the formula:

WNet = FNetdcosθ, where d is the distance moved by the box.

In this case, d = 71.0 m.

Therefore,

WNet = (264.36 N)(71.0 m)cos(70.9°) = 16,750 J

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Answer:0.216 kg

Explanation:

The volume of a sphere is given by the formula:

V = (4/3)πr^3

where r is the radius of the sphere. Since the diameter of the iron sphere is 6.6 cm, its radius is half of that or 3.3 cm (0.033 m).

The volume of the sphere is:

V = (4/3)π(0.033 m)^3

V = 2.76 x 10^-5 m^3

The density of iron is 7850 kg/m^3. We can use the density formula to find the mass of the sphere:

density = mass / volume

mass = density x volume

mass = 7850 kg/m^3 x 2.76 x 10^-5 m^3

mass = 0.216 kg

Therefore, the mass of the solid iron sphere is 0.216 kg.

Answer:

Density of iron = 7870 kg/m3 diameter of sphere = 3.00 cm radius of sphere r = 1.5 cm =0.015 m Volume of sphere V = (4/3) * pi* (r^3) = (4/3) x

Explanation:

Dopamine is key for : pleasure.  Dopamine lets you feel pleasure, satisfaction and also motivation. When one feels good that they have achieved something, then it is because one has surge of dopamine in the brain.

Dopamine is key for pleasure, reinforcement, and motivation and is commonly known as the "reward molecule". It is responsible for the feelings of pleasure and satisfaction that we experience after completing  rewarding activity, like eating food or engaging in social interaction.

Dopamine is also involved in reinforcement, which is the process by which brains learn to associate specific actions or behaviors with rewards.

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