Question 2) The power radiated by a lossless antenna is 10 W. The directional charact represented by the radiation intensity of U-11, cos' 0 [w the antenna are 0505, 0sps2x. Find: (Don't use approximate formulas) a) The value of B b) The half power beamwidth c) The first null beamwidth (4 marks [2.5 ma (2.5 m

The specific latent heat of fusion for the substance in question 1a is 80 kJ/kg, and the specific latent heat of vaporization for the substance in question 1b is 250 kJ/kg.

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A 90 kg person runs off a 10 m cliff into a body of water. What is their approximate gravitational potential energy after 1 second of free fall.  X A. 450J This is a miscalculation that results from not accounting for the acceleration due to gravity. X B.  The correct answer is C.

After 1 second of free fall, the person has fallen a distance of 4.9 meters (using the equation \(d = 1/2 * g * t^2\), where g is the acceleration due to gravity and t is the time). Therefore, the person is still 5.1 meters above the water. The gravitational potential energy (U) of the person can be calculated using the equation U = mgh, where m is the mass of the person, g is the acceleration due to gravity, and h is the height above the water. Substituting the values, we get:

\(U = (90 kg) * (9.8 m/s^2) * (5.1 m) = 4509 J\)

Therefore, the approximate gravitational potential energy of the person after 1 second of free fall is 4500 J.

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Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

Explanation:

Through -212 V potential difference must the electron pass to accomplish this

The difference in electric potential between two points, which is defined as the effort required per unit of charge to move a test charge between the two places, is known as voltage, electric potential difference, electric pressure, or electric tension.

The initial kinetic energy of the electron is,

\({K_i} = \frac{1}{2}mv_i^2\)

Here, m is the mass of the electron and

is the initial velocity of the electron.

\(\begin{array}{c}\\{K_i} = \frac{1}{2}\left( {9.11 \times {{10}^{ - 31}}\;{\rm{kg}}} \right){\left( {2.5 \times {{10}^6}\;{\rm{m/s}}} \right)^2}\\\\ = 2.88 \times {10^{ - 18}}\;{\rm{kg}}\\\end{array}\)

The final kinetic energy of the electron is,

\(\begin{array}{c}\\{K_f} = \frac{1}{2}\left( {9.11 \times {{10}^{ - 31}}\;{\rm{kg}}} \right){\left( {9 \times {{10}^6}\;{\rm{m/s}}} \right)^2}\\\\ = 3.68 \times {10^{ - 17}}\;{\rm{kg}}\\\end{array}\)

\(\begin{array}{c}\\{V_1} - {V_2} = \frac{{3.68 \times {{10}^{ - 17}}\;{\rm{J}} - 2.88 \times {{10}^{ - 18}}\;{\rm{J}}}}{{ - 1.6 \times {{10}^{ - 19}}\;{\rm{C}}}}\\\\ = - 212\;{\rm{V}}\\\end{array}\)

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The center of our galaxy is a much more crowded region than where the sun is found; we still see material falling toward the center and material has fallen in for billions of years

The Galactic Center (or Galactic Centre) is the rotational center, the barycenter, of the Milky Way galaxy. Its central massive object is a supermassive black hole of about 4 million solar masses, which is called Sagittarius A*, a compact radio source which is almost exactly at the galactic rotational center.

The Galactic Center is approximately 8 kiloparsecs (26,000 ly) away from Earth in the direction of the constellations Sagittarius, Ophiuchus, and Scorpius, where the Milky Way appears brightest, visually close to the Butterfly Cluster (M6) or the star Shaula, south to the Pipe Nebula.

There are around 10 million stars within one parsec of the Galactic Center, dominated by red giants, with a significant population of massive supergiants and Wolf–Rayet stars from star formation in the region around 1 million years ago. The core stars are a small part within the much wider galactic bulge.

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In the northern hemisphere, gyres rotate in the clockwise direction.

Therefore, the statement on question number 4 is false.

The water budget equation for the Creek drainage basin is: Precipitation = Stream discharge + Evapotranspiration ± Change in storage.

To develop the water budget equation for this problem, we need to account for the inputs and outputs of water in the Creek drainage basin. The water budget equation can be expressed as,

P = Q + ET ± ΔS

P = Precipitation input (mm)

Q = Stream discharge (m³/s)

ET = Evapotranspiration (mm)

ΔS = Change in storage (mm)

First, let's convert the units of the given values,

Precipitation input (P) = 550 mm

Stream discharge (Q) = 1.8 m³/s

Now, let's determine the evapotranspiration (ET) and change in storage (ΔS) terms. However, the problem doesn't provide information about ET, so we cannot calculate it accurately. The problem doesn't provide information about ΔS, so we cannot calculate it accurately.

Given these limitations, we can write the simplified water budget equation for this problem,

550 mm = 1.8 m³/s + ET ± ΔS

Please note that without information about evapotranspiration and changes in storage, we cannot fully determine the water budget for the Creek drainage basin.

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Answer:

.50x + 6

Explanation:

You dont know how many tickets she is gonna buy so you use x after the cost of each ticket, and since she needs $6 to get in you have to include that, i hope this helps

Answer:

Formular of kinteic energy is 0.5 × (mass x velocity)²

0.5 × 2500 kg × 35² =1,531,250 jules

The minimum and maximum braking distances needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection are as follows:

- For small cars: Minimum ≈ 1773.028 m, Maximum ≈ 1568.850 m

- For large trucks: Minimum ≈ 3285.760 m, Maximum ≈ 2409.595 m

To calculate the minimum and maximum braking distances, we can use the equations of motion for a decelerating vehicle.

The equation for the braking distance of a vehicle is given by:

d = (v^2) / (2 * μ * g)

where d is the braking distance, v is the initial velocity of the vehicle, μ is the coefficient of friction between the tires and the road, and g is the acceleration due to gravity.

Let's calculate the minimum and maximum braking distances separately for small cars and large trucks.

For small cars with a mass of 563 kg:

- Minimum braking distance:

  v = 8989 km/h = (8989 * 1000) / 3600 m/s = 2496.944 m/s

  μ_min = 0.536

  d_min = (v^2) / (2 * μ_min * g) = (2496.944^2) / (2 * 0.536 * 9.8) ≈ 1773.028 m

  Maximum braking distance:

  μ_max = 0.599

  \(d_max = (v^2) / (2 * μ_max * g) = (2496.944^2) / (2 * 0.599 * 9.8) ≈ 1568.850 m\)

For large trucks with a mass of 3951395 kg:

- Minimum braking distance:

  v = 8989 km/h = 2496.944 m/s (same as for small cars)

  μ_min = 0.350

 \(d_min = (v^2) / (2 * μ_min * g) = (2496.944^2) / (2 * 0.350 * 9.8) ≈ 3285.760 m\)

  Maximum braking distance:

  μ_max = 0.480

 \(d_max = (v^2) / (2 * μ_max * g) = (2496.944^2) / (2 * 0.480 * 9.8) ≈ 2409.595 m\)

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Answer:

divide the density of solution by density of water

EXPLANATION:

LIKE:

1.25÷1000kgm-3

Answer:

they are membrane proteins

Answer:

You've probably noticed that static electricity is more noticeable during the winter months. This is because the air is very dry. In the summer, the humidity and moisture in the air help electrons move more quickly, which makes it harder to build up a big static charge.

Explanation:

Answer:

A) the liquid particals move slower

Explanation:

vapor has a higher temp and have more energy than liquid so it's not c or d. and liquid has a lower boiling point than vapor

The rare disorder that includes the inability to feel pain is called Congenital analgesia. That is option B.

Congenital analgesia is defined as ,the name implies, a genetic disorder that rarely occurs which makes one not to be able to feel pain.

It is an autosomal recessive disorder in which there is genetic changes in the PMRD12 gene.

Therefore, the rare disorder that includes the inability to feel pain is called Congenital analgesia.

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Answer:

Explanation:

Mass = 50kg +7kg +4kg =61 kg

Velocity = 3m/s

Momentum = mass × velocity

\(p =61\times 3\\\\p =183MLT^-^1\)

Answer:

Mass= 61 kg. Success to The homework.

Answer: 2 PARAGRAPH ESSAY

Explanation:Flexibility and stretching: preventing sports injuries

  Many may know about all the different types of sports around the world. Lots can leave serious injuries. All across the world more than 1.2 million people have been injured because of sports or maybe more than that. Most of the injuries have been kids. To prevent these injuries from happening. One thing is stretching if you  don’t stretch your bones more than likely you are not into your flexibility and reflexus that for they are very weak. If there is not enough flexibility within your movement you are more than likely not able to stretch. It can help the growth within your bone as well. If you have growth in your bones then you will get less injuries. If you break a bone such as your leg or sprain there is a chance that you will have problems with that specific leg later on in life. To prevent these types of events from happening I say stretch before a sporting game. Stretching can help with the impact of your  flexibility. For example, Olympic gymnasts are able to do lots of tricks because their bones have flexibility in their bones. That's not saying that they will not hurt themselves, it's just such a low risk because their bone grew strength and flexibility. Some people's bones can be very sensitive then others so it might be kind of hard for them to be flexible. Another way to help people with their bone flexibility is drinking milk. Milk helps grow your bones. This can help prevent sport injuries in many ways. Lots of people get hurt in football to be more specific because it's such a durable sport as in tough. If they drink a lot of protein and milk before a game maybe even Carbohydrates their bones will be ready for the game. Flexibility and stretching can go lots of ways when it has to do with your body.  JUST STAY FIT

Answer:

The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. Groups are labeled at the top of each column.

Explanation:

Ethics in public administration promotes transparency, accountability, fairness, and responsible use of power, evolving to consider ethical dilemmas, stakeholder impact, and emerging challenges over time.

The importance of ethics in public administration lies in promoting transparency, accountability, fairness, and the responsible use of power to serve the public interest.

Over time, the understanding of ethics in public administration has evolved, with increased emphasis on professional standards, integrity, and the need to balance competing interests. It has shifted from a focus on rule-following to a more nuanced approach, considering ethical dilemmas and the impact of decisions on diverse stakeholders.

Additionally, ethical frameworks have been developed to guide decision-making, such as codes of conduct and ethics training programs. The changing landscape, including technological advancements and globalization, has also brought new ethical challenges, such as data privacy, social equity, and sustainability, requiring ongoing adaptation of ethical practices in public administration.

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In the DSM-5, each of the following has been assigned as an obsessive-compulsive-related disorder EXCEPT impulse-control disorder.

The Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition (DSM-5) includes a section on obsessive-compulsive and related disorders. The section contains eight different disorders, each with its own criteria.

In the DSM-5, each of the following has been assigned as an obsessive-compulsive-related disorder except for the impulse-control disorder.

Impulse-control disorder is not listed as an obsessive-compulsive-related disorder in DSM-5, and it is a separate condition. The DSM-5 classified Impulse-Control Disorder as an impulse-control disorder and not as an obsessive-compulsive-related disorder. It is an impulse control disorder characterized by an inability to resist the impulse, drive, or temptation to perform an act that is dangerous to oneself or others.In the DSM-5, the following are obsessive-compulsive-related disorders:

Obsessive-Compulsive Disorder (OCD)

Body Dysmorphic Disorder (BDD)

Trichotillomania (Hair-Pulling Disorder)

Excoriation (Skin-Picking) Disorder

Hoarding Disorder

Substance/Medication-Induced Obsessive-Compulsive

The DSM-5 is the fifth edition of the Diagnostic and Statistical Manual of Mental Disorders. It is a manual used by mental health professionals to diagnose mental illnesses. In DSM-5, each of the following has been assigned as an obsessive-compulsive-related disorder except impulse-control disorder. The DSM-5 classified impulse control disorder as an impulse control disorder and not as an obsessive-compulsive-related disorder.

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It takes 2.916s for the bicyclist to catch the runner and the runner had ran 17.2m with constant velocity of 3.5m/s

How quickly or slowly an object is moving can be determined by its velocity and speed. The need to determine which of two or more moving objects is moving faster arises frequently in our daily lives.

If both vehicles are traveling down the same road in the same direction, it is simple to determine which is moving more quickly.

To tell who is moving faster, though, is challenging if they are moving in the opposite direction. The idea of velocity is valuable in such circumstances

Lets say runner and bicyclist start at a point vi

The distance bicyclist cover to catch runner  is \(S = (v_i)t = \frac{1}{2} at^2\)

As the initial velocity is zero distance \(S = \frac{1}{2} at^2\)

For the runner, distance he could run before bicyclist overpassed him is

S = vt

Now as athletes was 7m ahead so

= (athlete)S  = (bicyclist)S

= (athlete)S = (bicyclist)S - 7

Substitute the values

= vt = \(\frac{1}{2} at^2\)

= 3.5t = \(\frac{1}{2} (2.4)t^2\)

= 3.5t = 1.2t²

= 3.5 = 1.2t

= t = 2.916s

Distance of runner = 3.5 × 2.916s

                                = 10.206 + 7m

                                = 17.2m

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The constant force that must be exerted on him to bring him to rest in a distance of 1.5 m in a time interval of 0.23 s opposite the fullback's direction of m is 0 N.

Given, distance = 1.5m and time = 0.23s.

Using the equation v = u + at, we can find acceleration.

Here, v = 0, u = 0 and t = 0.23 s, so acceleration a is equal to 0 m/s².

Using the equation F = ma, we can find force.

Here, m = 90 kg and a = 0 m/s², so force is equal to 0 N.

The negative sign with distance suggests that the direction of force is opposite to the direction of motion.

As we know that the direction of motion is opposite to the direction of fullback, so the constant force that must be exerted on him to bring him to rest in a distance of 1.5 m in a time interval of 0.23 s opposite the fullback's direction of m is 0 N.

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Temperature is not one of the basic properties  subatomic particle. The basic properties of a subatomic particle are its mass, spin, charge (electrical), and its interaction with other particles. Option C is Correct.

This means that regardless of how many interactions or reactions occur, the overall electric charge in a system never changes.

When subjected to an electromagnetic field, matter with an electric charge experiences a force. Objects with the same electric charge repel one another whereas those with differing charges attract one another. Electric charge can be positive or negative.

Subatomic particles like electrons and protons are capable of carrying electric charge, and the quantity of these particles in an object determines how much charge it has. Electric charge is conserved, which means that it can only be moved between objects and cannot be created or destroyed.

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When calculating building energy performance, these are included as process energy:

Ways to improve BEPS

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Answer:

When two objects collide, the forces that are exerted on each other are determined by their masses and speeds. If the mass of one of the objects is increased, the force that it exerts on the other object during the collision will also increase. Similarly, if the speed of one of the objects is increased, the force that it exerts on the other object during the collision will also increase. In general, the greater the mass and speed of an object, the greater the force that it will exert on another object during a collision. However, it is important to note that the direction of the forces will remain the same, regardless of the masses or speeds of the objects involved.

Answer:

nngh have

bjruh hjrhhj be rnrnnrnrnnnrnjrjnnnnnnnnrnrn n n and I was nrn

Explanation:

jbbbbbhhhjjnnnnnnnnnnjhvcc

Explanation:

Answer:

C- Frequency

Explanation:

Alpha particle loses 88.4% of its original kinetic energy in this elastic head-on collision with the gold nucleus.

The calculation to determine the percentage of kinetic energy lost by the alpha particle in this elastic head-on collision requires a long answer.
First, we can use the conservation of momentum to determine the final velocities of both the alpha particle and the gold nucleus after the collision. Since the gold nucleus is initially at rest, we have:

(mass of alpha particle) x (initial velocity of alpha particle) = (mass of alpha particle + mass of gold nucleus) x (final velocity of alpha particle)

Solving for the final velocity of the alpha particle, we get:

final velocity of alpha particle = (mass of alpha particle - mass of gold nucleus)/(mass of alpha particle + mass of gold nucleus) x (initial velocity of alpha particle)

Plugging in the values given in the question, we get:

final velocity of alpha particle = (4u - 197u)/(4u + 197u) x (initial velocity of alpha particle)

final velocity of alpha particle = -0.986 x (initial velocity of alpha particle)

This negative sign indicates that the alpha particle moves in the opposite direction after the collision.

Next, we can use the conservation of energy to determine the percentage of kinetic energy lost by the alpha particle in the collision. Since the collision is elastic, the total kinetic energy before and after the collision should be the same. Therefore:

(initial kinetic energy of alpha particle) + (initial kinetic energy of gold nucleus) = (final kinetic energy of alpha particle) + (final kinetic energy of gold nucleus)

The initial kinetic energy of the gold nucleus is zero since it is initially at rest. The kinetic energy of a particle is given by:

kinetic energy = (1/2) x (mass of particle) x (velocity of particle)^2

Plugging in the values for the alpha particle before and after the collision, we get:

(initial kinetic energy of alpha particle) = (1/2) x (4u) x (initial velocity of alpha particle)^2

(final kinetic energy of alpha particle) = (1/2) x (4u) x (final velocity of alpha particle)^2

Using the equations we derived earlier for the final velocity of the alpha particle, we can simplify this to:

(final kinetic energy of alpha particle) = (1/2) x (4u) x (0.986 x initial velocity of alpha particle)^2

(final kinetic energy of alpha particle) = (0.484 x initial kinetic energy of alpha particle)

Therefore, the percentage of kinetic energy lost by the alpha particle in the collision is:

percentage of kinetic energy lost = [(initial kinetic energy of alpha particle) - (final kinetic energy of alpha particle)]/(initial kinetic energy of alpha particle) x 100%

percentage of kinetic energy lost = [(1/2) x (4u) x (initial velocity of alpha particle)^2 - (0.484 x (1/2) x (4u) x (initial velocity of alpha particle)^2)]/[(1/2) x (4u) x (initial velocity of alpha particle)^2] x 100%

percentage of kinetic energy lost = 88.4%

Therefore, the alpha particle loses 88.4% of its original kinetic energy in this elastic head-on collision with the gold nucleus.

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