An object of mass 4kg moves in a circle at radius 8cm at uniform speed of 32m/s calculate (A)The angular velocity (B) The centripetal force

Answer:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

Explanation:

Answer:

A) but be sure and read the answer.

Explanation:

If the object does not move at all, (that's an important restriction) the net force = 0. That being so, the acceleration must be 0 as well. But there is no law saying that there cannot be a constant motion and that's why the restriction is important.

A falling object accelerates from -10.00 to-30.00m/s, it takes approximately 2.04 seconds for the falling object to accelerate from -10.00 m/s to -30.00 m/s.

To determine the time it takes for a falling object to accelerate from one velocity to another:

v = u + at

Here, it is given that:

Initial velocity (u) = -10.00 m/s

Final velocity (v) = -30.00 m/s

Since the object is falling, we can assume that the acceleration (a) is due to gravity and is approximately -9.8 m/s² (negative because it's directed downwards).

Substituting the values into the equation:

-30.00 = -10.00 + (-9.8)t

Simplifying the equation:

-30.00 + 10.00 = -9.8t

-20.00 = -9.8t

Dividing both sides by -9.8:

t = -20.00 ÷ -9.8

t ≈ 2.04 seconds

Thus, it takes approximately 2.04 seconds for the falling object to accelerate from -10.00 m/s to -30.00 m/s.

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Friction is not always negligible, as it can have significant effects on the behavior of objects in motion.

Friction is a force that opposes the motion of two surfaces in contact with each other. The magnitude of the frictional force depends on the nature of the surfaces in contact, the force pressing the surfaces together, and the relative motion between the surfaces.

In some cases, the effects of friction can be ignored because they are small compared to other forces at play. For example, in situations where an object is moving through a fluid, such as air or water, the effects of friction may be negligible compared to the forces of drag and buoyancy.

In other situations, the effects of friction cannot be ignored. For example, in a car's braking system, the friction between the brake pads and the rotor is necessary to slow down and stop the car. Without this friction, the car would not be able to come to a complete stop.

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The atomic radius of copper is approximately 128 picometers (pm).

The atomic radius of an element is defined as half the distance between the nuclei of two identical adjacent atoms in a molecule. In the case of metallic copper, the copper atoms are arranged in a crystal lattice, and the distance between two adjacent copper atoms in the lattice is known as the interatomic distance or lattice parameter.

We are given that the distance between two adjacent copper atoms in metallic copper is 256 pm. Since this is the distance between the nuclei of two adjacent atoms, the sum of the atomic radii of the two copper atoms is equal to 256 pm.

Therefore, the atomic radius of copper can be calculated as follows:

Atomic radius of Cu = (interatomic distance between adjacent Cu atoms) / 2

Atomic radius of Cu = 256 pm / 2

Atomic radius of Cu = 128 pm

Hence, the atomic radius of copper is approximately 128 picometers (pm).

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Answer:

the exact center of its mass

The distance travelled by Arthur is 7 km and his displacement is 5 km.

The distance travelled by Arthur during the entire motion is determined by summing the entire path covered during the motion.

Distance = 3 km + 4 km

Distance = 7 km

The displacement of Arthur during the entire motion is obtained by calculating the length of the shortest path between the initial and final position.

d = √(a² + b)

where;

d = √(3² + 4²)

d = 5 km

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Answer:

the answer to this would be D.

Answer: Its D but it might swap to another letter so keep your eye out bud pass this test good luck

Explanation:

Static electricity represents charges without movement, while electric current represents charges in motion (electric flow).

The fuse is a component of electrical installations that is interrupted or melts when the current is excessive. Fuses are composed of a sheet or filament made of an alloy or metal that is characterized by a low melting point. This element is located at a strategic point in the electrical installation so that it melts if the intensity of the current exceeds a certain value. Thus, the fuse interrupts the current and safeguards the integrity of the conductors, minimizing the risk of fire or breakdown.

Answer

30 mi/gal * (1.61 km / mi ) /  (3.78 L/gal)   = 30 * .426 km/L = 12.8 km/L

Given,

The image distance, v=17.22 cm

The focal length of the lens, f=3.05 cm

From the thin lens formula,

Where u is the object distance.

On rearranging the above equation,

On substituting the known values,

Thus the object distance should be 3.71 cm

To determine the reject rate (RR) for each scenario, we need to consider the capability indices Cp and Cpk along with the specifications for tensile strength. The reject rate represents the proportion of parts that do not meet the specifications.

1. RR = 0.27%

2. RR = 16.03%

3. RR = 0%

4. RR = 0.003%

5. RR = 0.27%

6. RR = 2.28%

7. RR = 29.93%

8. RR = 0%

9. RR = 5.87%

Please note that these calculations assume a normal distribution of tensile strength and that the process is in statistical control. The reject rate is obtained by evaluating the proportion of values falling outside the specified limits based on the process capability indices Cp and Cpk.

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Title: Kinetic Energy Lab Report

Abstract:

The Kinetic Energy Lab aimed to investigate the relationship between an object's mass and its kinetic energy. The experiment involved measuring the mass of different objects and calculating their respective kinetic energies using the formula KE = 0.5 * mass * velocity^2. The velocities of the objects were kept constant throughout the experiment. The results showed a clear correlation between mass and kinetic energy, confirming the theoretical understanding that kinetic energy is directly proportional to an object's mass.

Introduction:

The concept of kinetic energy is an essential aspect of physics, describing the energy possessed by an object due to its motion. According to the kinetic energy equation, the amount of kinetic energy depends on both the mass and velocity of the object. This experiment focused on exploring the relationship between an object's mass and its kinetic energy, keeping velocity constant. The objective was to determine if an increase in mass would result in a corresponding increase in kinetic energy.

Methodology:

1. Gathered various objects of different masses.

2. Measured and recorded the mass of each object using a calibrated balance.

3. Kept the velocity constant by using a consistent method to impart motion to the objects.

4. Calculated the kinetic energy of each object using the formula KE = 0.5 * mass * velocity^2.

5. Recorded the calculated kinetic energies for each object.

Results:

The data collected from the experiment is presented in Table 1 below.

Table 1: Mass and Kinetic Energy of Objects

Object    Mass (kg)   Kinetic Energy (J)

----------------------------------------

Object A   0.5        10.0

Object B   1.0        20.0

Object C   1.5        30.0

Object D   2.0        40.0

Discussion:

The results clearly demonstrate a direct relationship between mass and kinetic energy. As the mass of the objects increased, the kinetic energy also increased proportionally. This aligns with the theoretical understanding that kinetic energy is directly proportional to an object's mass. The experiment's findings support the equation KE = 0.5 * mass * velocity^2, where mass plays a crucial role in determining the amount of kinetic energy an object possesses. The constant velocity ensured that any observed differences in kinetic energy were solely due to variations in mass.

Conclusion:

The Kinetic Energy Lab successfully confirmed the relationship between an object's mass and its kinetic energy. The data collected and analyzed demonstrated that an increase in mass led to a corresponding increase in kinetic energy, while keeping velocity constant. The experiment's findings support the theoretical understanding of kinetic energy and provide a practical example of its application. This knowledge contributes to a deeper comprehension of energy and motion in the field of physics.

References:

[Include any references or sources used in the lab report, such as textbooks or scientific articles.]

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Question :-

Answer :-

Explanation :-

As per the provided information in the given question, The Resistance is given as 20 Ohm's . Current is given as 1.5 Amperes . And, we have been asked to calculate the Voltage .

For calculating the Voltage , we will use the Formula :-

\( \bigstar \: \: \: \boxed {\sf { \: Voltage \: = \: Current \: \times \: Resistance \: }} \)

Therefore , by Substituting the given values in the above Formula :-

\( \Longrightarrow \: \: \: \sf { Voltage \: = \: Current \: \times \: Resistance } \)

\( \Longrightarrow \: \: \: \sf { Voltage \: = \: 1.5 \: \times \: 20 } \)

\( \Longrightarrow \: \: \: \bf { Voltage \: = \: 30 } \)

Hence :-

\( \underline {\rule {180pt}{4pt}} \)

Additional Information :-

\(\Longrightarrow \: \: \: \sf {Voltage \: = \: Current \: \times \: Resistance} \)

\( \Longrightarrow \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance}} \)

\( \Longrightarrow \: \: \: \sf {Resistance \: = \: \dfrac {Voltage}{Current} } \)

Answer:

Explanation:

Given:

To Find:

Solution:

Using formula:

By Substituting the required values,

⇢ Voltage = 1.5 × 20

⇢ Voltage = 30 Volts.

Hence,

Answer:

15

Explanation:

225/(165-150)=15g/cm3

Answer:

They include;

1. Earthquakes

2. Floods

3. Hurricane

4. Landslide

5. Tsunami

6. Tornadoes

Explanation:

Emergencies are sudden disasters that occur occasionally and unexpectedly, resulting in massive destruction of lives and property, and which require prompt actions to resolve. When some of these emergency situations occur, the electric grids could fail, resulting in a power outage in the affected town or community. Rescue workers might then find communication difficult because they cannot get energy to their electrical devices.

Examples of emergency situations that could have this type of effect include; Earthquakes, Floods, Hurricane, Landslide, Tsunami, and Tornadoes.

The weight of a \(45 kg\) box is \(441.45 N\).

Weight refers to the measure of the force exerted on an object due to the gravitational pull of the Earth.

Given the mass of the box is \(m=45 kg\).

The weight of an object (\(W\)) can be found by multiplying the mass of the object (\(m\)) by the acceleration due to gravity (\(g\)).

So, \(W=mg\).

It is known that acceleration due to gravity, \(g=9.81 m/s^2\).

Hence, the weight of the box, \(W=(45kg)\times (9.81 m/s^2)\).

\(\Rightarrow W=441.45 (kg\cdot m/s^2)\\\Rightarrow W=441.45 N\)

Therefore, the weight of the box is \(441.45 N\).

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Answer:

One gram atom of any substance contains one Avogadro number of atoms. Therefore, the number of atoms of U-235 present in 235 g of the substance = Avogadro Number = 6.02×10^23 atoms

The number of U-235 atoms present in 1 g of the substance = (6.02×10^23/235) atoms=2.56×10^21 atoms

Energy released in one atom fissioning = 200 MeV= 200 MeV × 1.6 × 10^-6 erg/MeV= 3.2 ×10^-4 erg/atom.

Energy released in the fission of all the atoms contained in 1g of Uranium-235 fissioning = (2.56×10^21 atom)×(3.2×10^-4 erg/atom)=8.19 ×10^17 erg

Now

1 kWh = 1000 Watt hour = 1000×( 10^7 erg/s)×3600s = 3.6 ×10^13 erg/kWh

So energy in kWh released in fission of 1 g of U-235 = (8.19× 10^17 erg)÷(3.6×10^13 erg/kWh) = 2.55 ×10⁴ kWh.

Added on 19th February, 2019.

The energy consumption of an average home in India in a month is between 300–500 kWh in the metropolitan cities. Let us take it at 400 kWh. So one gram of nuclear fuel (uranium) would be able to meet the energy requirements of a town of about 6000 for a month.

The energy in a fissile material is highly concentrated, compared to other types of fuels like wood, coal, diesel etc. So nuclear powered submarines can remain below water for very long periods of time, as they do not need to surface for getting fuel.

Explanation:

Energy in kWh released in fission of 1 g of U-235 is 2.55 ×10⁴ kWh.

One gram atom of any substance contains one Avogadro number of atoms. Therefore, the number of atoms of U-235 present in 235 g of the substance = Avogadro Number = 6.02×10^23 atoms

The number of U-235 atoms present in 1 g of the substance = (6.02×10^23/235) atoms=2.56×10^21 atoms

Energy released in one atom fissioning = 200 MeV= 200 MeV × 1.6 × \(10^{-6}\) erg/MeV= 3.2 ×\(10^{-4}\) erg/atom.

The energy released in the fission of all the atoms contained in 1g of Uranium-235 fissioning = (2.56×\(10^{21}\) atom)×(3.2×\(10^{-4}\) erg/atom) = 8.19 ×\(10^{17}\) erg

1000 Watt hour = 1000×( \(10^{7}\) erg/s)×3600s = 3.6 ×10^13 erg/kWh

So the energy in kWh released in fission of 1 g of U-235 = 2.55 ×10⁴ kWh.

Fission takes place when a neutron slams into a bigger atom, forcing it to excite and break up into smaller atoms—additionally referred to as fission products. Extra neutrons are also launched that may initiate a sequence reaction. whilst each atom splits, a high-quality amount of electricity is released.

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The individual effect of salinity reduced organism performance in 43% of the observations, including decreased survival and growth, increased osmolyte content in bodily fluids, altered metabolic rates, etc.

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The magnitude of the vector displacement is 15.65 m.

The resultant force acting on the object is calculated as follows;

\(F = \sqrt{F_x^2 + F_y^2} \\\\F = \sqrt{4^2 + 6^2} \\\\F = 7.21 \ N\)

The displacement of the vector is calculated as follows;

W = Fs cosθ

\(s = \frac{W}{Fcos(\theta)} \\\\s = \frac{106}{7.12 \times cos(18)} \\\\s = 15.65 \ m\)

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Answer:

Part A

1) At the starting point, we have;

PE = 40,000 J

2) PE = 0 J, KE = 40,000 J

3) KE = 20,000 J

4) PE = 15,000 J

5) KE = 32,500 J

6) KE = 40,000 J, PE = 0 J

7) KE = 35,000 J

8) KE = 40,000 J, PE = 0 J

Part B

The total Mechanical Energy = ME = 40,000 J

At the final point, we have;

ME = KE + PE = 40,000 J + 0 J = 40,000 J

Explanation:

Part A

By the law of conservation of energy, we have;

ME = PE + KE

Where;

ME = The total Mechanical Energy of the system

PE = The Potential Energy of the system

KE = The Kinetic Energy of the system

Where there is no friction, we have;

At the final stage, KE = 40,000 J. PE = 0 J

Therefore, ME = PE + KE = 40,000 J + 0 J = 40,000 J

1) At the starting point, we have;

KE = 0 J, therefore, PE = ME - KE = 40,000 J - 0 J = 40,000 J

2) At the bottom of the roller coaster, at the same level the PE is taken as PE = 0 J at the final stage, we have;

PE = 0 J, therefore, KE = ME - PE = 40,000 J - 0 J = 40,000 J

3) Where PE = 20,000 J, KE = ME - PE = 40,000 J - 20,000 J = 20,000 J

4) Where KE = 25,000 J, PE = ME - KE = 40,000 J - 25,000 J = 15,000 J

5) Where PE = 7,500 J, KE = ME - PE = 40,000 J - 7,500 J = 32,500 J

6) At the bottom KE = 40,000 J, PE = 0 J

7) Where PE = 5,000 J, KE = ME - PE = 40,000 J - 5,000 J = 35,000 J

8) KE = 40,000 J, PE = 0 J

Part B

The given that there is no friction nor air resistance, the total Mechanical Energy, ME, is constant and equal to the sum of the Potential Energy, PE and the Kinetic Energy, KE, as follows;

ME = KE + PE

At the final point, we have;

ME = 40,000 J + 0 J = 40,000 J

The total Mechanical Energy = ME = 40,000 J

A truck accelerating at 0.0083 meters/second2 covers a distance of 5.8 ×\(10^4\) meters. If the truck's mass is 7,000 kilograms, the work done to reach this distance is 3.3× \(10^6\) joules.

The correct answer is option E.

To calculate the work done by the truck to cover a distance of 5.8 × \(10^4\)meters, we need to use the equation for work:

Work = Force × Distance

In this case, the force can be calculated using Newton's second law:

Force = mass × acceleration

Where:

Acceleration (a) = \(0.0083 meters/second^2\)

Distance (d) = 5.8 ×\(10^4\) meters

Mass (m) = 7,000 kilograms

First, let's calculate the force exerted by the truck:

Force = mass × acceleration = (7,000 kg) ×\((0.0083 meters/second^2)\)= 57.1 Newtons

Next, we can calculate the work done:

Work = Force × Distance = (57.1 N) × (5.8 × 10^4 meters) = 3.3158 × \(10^6\)joules

Rounded to the nearest significant figure, the work done by the truck is approximately 3.3 × \(10^6\) joules.

Therefore, the correct answer is E.

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The question probable may be:

A truck accelerating at 0.0083 meters/second2 covers a distance of 5.8 ×\(10^4\)meters. If the truck's mass is 7,000 kilograms, what is the work done to reach this distance?

A. 1.7 × \(10^6\) joules

B. 3.4 ×\(10^6\) joules

C. 5.6 × \(10^6\)joules

D. 6.8 ×  \(10^6\)joules

E. 3,3  ×\(10^6\)  joules

On a calm, sunny, summer day, the air in a very thin layer in contact with the ground can approach 140 degrees Fahrenheit after being heated primarily by conduction.

Conduction refers to the transfer of heat between objects that are in contact with each other. On a calm, sunny summer day, the air in a very thin layer in contact with the ground can approach 140 degrees Fahrenheit after being heated primarily by conduction. This is because the sun's rays heat up the ground, and then the heat is transferred to the air in contact with the ground through conduction.

As a result, the air in contact with the ground gets heated up to temperatures that can be much higher than the air just a few feet above it. This is why it's always a good idea to wear shoes when walking on hot pavement or sand, as the ground can get hot enough to cause burns.

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A: The  momentum of the first player be 450  kg.m/s.

The  momentum of the second player be 800 kg.m/s.

B:  The total momentum of the football players be 1250  kg.m/s.

C: The velocity of the football players after the collision be 7.14 m/s.

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

Given that:

One has a mass of 75 kg and is moving at 6 m/s.

The second has a mass of 100 kg and is moving at 8 m/s

So, momentum of the first player = mass * velocity

= 75×6 kg.m/s

= 450  kg.m/s.

The momentum of the second player = mass * velocity

= 100×8  kg.m/s

= 800 kg.m/s.

If the two football players collide, the total momentum of the football players be =  450  kg.m/s + 800 kg.m/s = 1250  kg.m/s.

The velocity of the football players after the collision be = total momentum/total mass =  1250  kg.m/s./(75+100)kg = 7.14 m/s.

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Answer: It has the skill to nest

sorry if I'm wrong

Answer:

B: A colorful beak

Explanation:

i hope this helped!

Answer:

Acceleration → Change in velocity divided by time.

Proportional → Net force and acceleration

Inversely Proportional → Mass and acceleration

Net force Not Equal to zero → Driving a car up the on-ramp of the freeway

Net force Equal to zero → Driving a car with cruise control on, straight down.

Acceleration → Net force divided by mass

Force → Mass times acceleration.

Explanation:

Below we will explain each choice.

A) Acceleration by means of the following expression of kinematics is defined as the change of velocities in a given time.

\(a=\frac{v_{f}-v_{o}}{t}\)

where:

a = acceleration [m/s²]

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

t = time [s]

B)

Newton's Second Law tells us that the sum of forces in a body is equal to the product of mass by acceleration so we have the following expression.

F = m*a

where:

F = force [N] (units of Newtons)

m = mass [kg]

a = acceleration [m/s²]

As we can see in the previous equation The Force is proportional to the acceleration, that is to say, every time the acceleration increases or decreases this same will happen with the force, increases or decreases.

C)

In Newton's second law we see an inversely proportional relationship, which is best seen if we clear the term of acceleration.

\(a=\frac{F}{m}\)

In any mathematical expression where a term is in the denominator, we can say that this is inversely proportional to the term to the left of the equal sign.

D)

When a vehicle climbs a hill or climbs a slope, forces such as the weight of the vehicle act in the opposite direction to the movement, so the sum of the forces cannot be equal to zero.

E)

When a vehicle moves on a track at a constant speed that is to say at cruising speed, we say that its acceleration is zero, therefore the sum of the forces acting on the moving vehicle is equal to zero.

F)

Acceleration is defined as the relationship between Force and mass

\(a =\frac{F}{m}\)

G)

Force is defined as the product of mass by acceleration.

F = m*a