What is the velocity of a 1,000.0 kg car if its kinetic energy is 200 kJ?

Answer:

Explanation:

We can solve this problem using trigonometry. Let's draw a diagram:

          |\

          | \

          |  \   <- Sun rays

          |   \

          |    \

          |     \

          |      \

          ---------

           Pool

The angle between the sun rays and the horizontal line is 90 - 25 = 65 degrees. Let's call the depth of the pool "d". We want to find the value of "d" that makes the bottom of the pool completely shaded.

We can see that the triangle formed by the sun rays, the top edge of the pool, and the bottom edge of the shaded area is a right triangle. The angle between the sun rays and the top edge of the pool is also 65 degrees, because the top edge is parallel to the ground.

Using trigonometry, we can write:

tan(65 degrees) = d / 4.9 m

Solving for "d", we get:

d = 4.9 m * tan(65 degrees)

Using a calculator, we find:

d ≈ 13.7 m

Therefore, the pool is approximately 13.7 meters deep.

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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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:

1.3 m/s

Explanation:

Answer:White Candle

Explanation:because it reflects all colors of light.

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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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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Dividing this by 24.0 s gives a power output of 23,600 W.

Power is the ability to influence or control the behaviour of people, events, or things. It is the capacity to make decisions and take action in order to effect change. Power can be physical, economic, political, cultural, or social, and can be held by individuals, organizations, or governments. Power can be used positively to create positive change, or negatively to oppress or exploit. It can be used to create a sense of safety, stability, and justice, or to create uncertainty and chaos.

The power output of the crane is calculated by dividing the work done by the crane (the force multiplied by the distance moved) by the time taken to do the work. In this case, the work done is 13,700 N x 41.4 m = 570,380 Nm. Dividing this by 24.0 s gives a power output of 23,600 W.

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Answer:25 c

Explanation:

The average speed of the skier is 53.3 m/min.

option C.

Average speed is a measure of how fast an object or a person travel over a certain distance in a specific amount of time.

It is calculated by dividing the total distance traveled by the time it took to travel that distance.

The formula for average speed is:

Average Speed = Total Distance Traveled / Time Taken

The average speed of the skier is calculated as follows;

Average speed = ( 160 m ) / 3 mins

Average speed = 53.3 m/min

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Answer:

Look down below

Explanation:

300Hz is the 5th harmonic in a 60 Hz system, or the 6th harmonic in a 50 Hz system.

While the three stable isotopes of Argon share the same number of protons and electrons, they differ in their atomic masses due to varying numbers of neutrons.

This means they have identical electron configurations and chemical properties. Additionally, they all have the same number of electrons, 18, which determines their chemical behavior and bonding capabilities.

However, the isotopes differ in their neutron numbers, which gives rise to their atomic masses. Argon-36 has 18 protons and 18 neutrons, Argon-38 has 18 protons and 20 neutrons, and Argon-40 has 18 protons and 22 neutrons.

This variation in neutron count leads to differences in atomic mass. Consequently, the isotopes exhibit different atomic weights, with Argon-36 having a mass of approximately 36 atomic mass units (amu), Argon-38 around 38 amu, and Argon-40 approximately 40 amu.The differing atomic masses affect the isotopes' physical properties.

For instance, the isotopes may have different boiling points, melting points, or densities due to variations in the average mass of their atoms. These subtle differences are relevant in scientific research, especially in fields like geochronology and radiometric dating, where the abundance ratios of these isotopes are utilized to determine the age of rocks and minerals.

In summary,  These differences give rise to variations in their atomic weights and physical properties.

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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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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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The interia is varied and it depends on the mass, itt will be sameshape and size so it will be light closer.

What is volume ?

Space is occupied by every three-dimensional object. Its volume serves as a gauge for this area. The area contained by an object's limits in three-dimensional space is referred to as its volume. Another name for it is an object's capacity.

What is element ?

The atoms make the element. The atoms are created up of the element; it is a pure substance it can't be able to split up. The element is created up of a single kind of atom. Atoms are the main thing for the creation of elements. Elements are classified according to their performance and physical properties. Elements are placed based on the performance in the Mendeleev periodic table.

Therefore, The interia is varied and it depends on the mass, itt will be sameshape and size so it will be light closer.

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The force required to lift a load of 100 newtons using machine with mechanical advantage of 4 is 25 newtons.

The ratio of output force to the input force is called as mechanical advantage of a machine. In this case, mechanical advantage is given as 4.

As, Mechanical Advantage = Output Force / Input Force

4 = Output Force / Input Force

So, Output Force = Mechanical Advantage x Input Force

Output Force = 4 x Input Force

100 newtons = 4 x Input Force

Input Force = 100 newtons / 4

Input Force = 25 newtons

Therefore, the force required to lift a load of 100 newtons using a machine with a mechanical advantage of 4 is 25 newtons.

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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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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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(a)The rifle exerts a force of 628 N on the bullet while it is in the barrel.

(b) (i)  There is no friction between the bullet and the barrel, there is no force of friction to consider.

    (ii) Just after it has left the barrel, the only force acting on the bullet would be the force of gravity pulling it downward.

(c) The rifle gives the bullet an acceleration of about 24885 g's.

(d) the bullet is in the barrel for 0.030 s.

What is Force?

Force is a physical quantity that describes the interaction between two objects, causing a change in motion. Specifically, it is an influence that can cause an object to accelerate, change direction, or deform. Force is defined as the product of mass and acceleration, according to Newton's second law of motion. Mathematically, force can be represented by the equation:

                              F = m*a

(a) To determine the force that the rifle exerts on the bullet while it is in the barrel, we need to use the equation for the work done by a constant force:

                                   W = Fd

where W is the work done, F is the force, and d is the distance over which the force is applied. We can rearrange this equation to solve for the force:

F = W/d

We know that the work done on the bullet is equal to the change in kinetic energy:

W = ΔK = (1/2)mv^2

where m is the mass of the bullet and v is its velocity. Plugging in the given values, we get:

W = (1/2)(0.00420 kg)(370 m/s)^2 = 283 J

The distance over which the force is applied is given as 45.0 cm = 0.45 m. So the force exerted by the rifle on the bullet is:

F = W/d = 283 J / 0.45 m = 628 N

Therefore, the rifle exerts a force of 628 N on the bullet while it is in the barrel.

(b) (i) The free-body diagram of the bullet while it is in the barrel would show two forces acting on it: the force of the rifle pushing it forward, and the force of gravity pulling it downward. Since there is no friction between the bullet and the barrel, there is no force of friction to consider.

(ii) Just after it has left the barrel, the only force acting on the bullet would be the force of gravity pulling it downward.

(c) We can calculate the acceleration of the bullet using the formula:

a = Δv/t

where Δv is the change in velocity and t is the time for which the acceleration occurs. We know that the initial velocity of the bullet is 370 m/s and that it starts from rest, so the change in velocity is:

Δv = 370 m/s

The distance over which the acceleration occurs is given as 45.0 cm = 0.45 m. Using the formula for distance traveled with constant acceleration, we can find the time for which the acceleration occurs:

d = (1/2)at^2

0.45 m = (1/2)a(t^2)

Solving for t, we get:

t = sqrt(0.9/a)

Plugging this into the equation for acceleration, we get:

a = Δv/t = 370 m/s / sqrt(0.9/a)

Solving for a, we get:

a = 2.44 x 10^5 m/s^2

To express this acceleration in units of g's, we can divide by the acceleration due to gravity:

a/g = 2.44 x 10^5 m/s^2 / 9.81 m/s^2 = 24885

Therefore, the rifle gives the bullet an acceleration of about 24885 g's.

(d) Using the same formula as before for distance traveled with constant acceleration, we can solve for the time the bullet is in the barrel:

d = (1/2)at^2

0.45 m = (1/2)(2.44 x 10^5 m/s^2)t^2

Solving for t, we get:

t = [tex]\sqrt{(0.00092)}[/tex]s = 0.030 s

Therefore, the bullet is in the barrel for 0.030 s.

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The bullet is in the barrel for 7.05 x 10^-5 seconds. Acceleration can also be caused by changes in direction, such as when an object moves in a circular path.

What is Acceleration?

Acceleration is a fundamental concept in physics and plays an important role in understanding the motion of objects. It is used to describe the behavior of objects in a wide range of applications, from simple everyday situations such as cars accelerating and braking, to more complex phenomena such as the acceleration of particles in particle accelerators or the acceleration of celestial bodies in space.

(a) To find the force exerted by the rifle on the bullet, we can use the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (muzzle velocity), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance traveled (45.0 cm = 0.45 m).

Rearranging this equation to solve for acceleration:

a = (v^2 - u^2) / 2s

Plugging in the given values:

a = (1200 m/s)^2 / (2 x 0.45 m) = 3.20 x 10^6 m/s^2

The force exerted by the rifle on the bullet can be found using Newton's second law:

F = ma

where F is the force, m is the mass of the bullet (4.20 g = 0.00420 kg), and a is the acceleration we just calculated:

F = 0.00420 kg x 3.20 x 10^6 m/s^2 = 13,440 N

Therefore, the rifle exerts a force of 13,440 N on the bullet while it is in the barrel.

(b)

(i) Free-body diagram of the bullet while it is in the barrel:

The only force acting on the bullet while it is in the barrel is the force exerted by the rifle, which is directed to the right.

  |

  |

-->|  F

  |

  |

Once the bullet has left the barrel, it is subject to air resistance, which we will assume acts in the opposite direction to the velocity of the bullet. The force of gravity on the bullet is negligible for this problem.

  |

  |

<--|  F_air

  |

  |

(c) The acceleration given to the bullet can be expressed in terms of g's by dividing by the acceleration due to gravity, g:

a_g = a / g = (3.20 x 10^6 m/s^2) / 9.81 m/s^2 = 326,000 g's

Therefore, the rifle gives the bullet an acceleration of 326,000 g's.

(d) The time the bullet is in the barrel can be found using the kinematic equation:

s = ut + (1/2)at^2

where s is the distance traveled (0.45 m), u is the initial velocity (0 m/s), a is the acceleration we calculated earlier (3.20 x 10^6 m/s^2), and t is the time the bullet is in the barrel (which we want to find).

Rearranging and solving for t:

t = sqrt(2s/a) = sqrt(2 x 0.45 m / 3.20 x 10^6 m/s^2) = 7.05 x 10^-5 s

Therefore, the bullet is in the barrel for 7.05 x 10^-5 seconds.

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If a 40 N block is resting on a rough horizontal table with a coefficient of static friction is 12 N.

Static friction is a force that resists the motion of two objects that are in contact with each other. It is caused by the forces of attraction between the two objects and is usually greater than the force of kinetic friction. The forces of static friction oppose the movement of the two objects and can be overcome by applying a force greater than the static friction.

The maximum force the block can withstand before it starts to move is 40 N multiplied by the coefficient of static friction.

The coefficient of static friction between the block and the table is determined by the materials of the block and the table and the surface roughness of the table.

If the coefficient of static friction is 0.3, then the maximum force the block can withstand before it starts to move is 40 N × 0.3 = 12 N.

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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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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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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 kinship chart of family is given in the image attached. The kinship chart of the Kardashian-Jenner family with "Kris Jenner" as "ego":

In this kinship chart, Kris Jenner is the "ego" or the main person we are focusing on. Her children are Kourtney Kardashian (daughter) and Robert Kardashian (son). Kourtney has three children: Mason, Penelope, and Reign. Robert has four children: North, Saint, Chicago, and Psalm.

Therefore, Each arrow represents a biological relationship, and each symbol represents a gender and a generational difference. "F" stands for female, "M" stands for male, and "-" stands for unknown gender.

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We can use the kinematic equations to solve for the time it takes for the soccer ball to hit the ground.

Let's use the following variables:

h for the initial height of the soccer ball (12 m)

vi for the initial velocity of the soccer ball (4 m/s)

vf for the final velocity of the soccer ball (0 m/s)

d for the horizontal distance the soccer ball travels before hitting the ground (6.3 m)

t for the time it takes for the soccer ball to hit the ground (what we want to find)

Using the kinematic equation for vertical motion, we can find the time it takes for the soccer ball to reach the ground:

vf^2 = vi^2 + 2*a*h

0 = 4^2 + 2*(-9.8)*12

0 = 16 - 235.2

t_vertical = sqrt(235.2/16) = 3.05 s

So it takes 3.05 seconds for the soccer ball to hit the ground if we consider only the vertical motion.

Next, using the kinematic equation for horizontal motion, we can find the time it takes for the soccer ball to travel the horizontal distance of 6.3 m:

d = vi*t_horizontal

t_horizontal = d/vi = 6.3/4 = 1.575 s

So it takes 1.575 seconds for the soccer ball to travel the horizontal distance of 6.3 m.

Since we know the soccer ball takes the same amount of time to travel horizontally as it does to hit the ground, we can conclude that the time it takes for the soccer ball to hit the ground is 1.575 seconds.

Answer:

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

We know that I=Q/T

Q=25×4.15=

112.5c

Explanation:

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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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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(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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