How are magnetic fields like vectors?

The power to stop the car with kinetic energy of a car is  [tex]8*10^{6} J[/tex] as it travels along a horizontal road is  [tex]8*10^{5} watt[/tex], option B

Kinetic energy can be seen as one that is been recorded when an object is able to move from a place , in a broad term we can say this is the energy that can be attributed to that of someone leaving a place and go to another place hence we can see it as the one in the motion.

The definition of energy as the "power to accomplish work" refers to the capacity to apply a force that moves an object. Even if the word is vague, it is clear what energy actually means: it is the force that causes objects to move. The two types  can be attributed to the one we know which are kinetic and potential energy.

[tex]Power \frac{Energy}{time}[/tex]

[tex]Energy = 8*10^{6} J[/tex]

[tex]time = 10 s[/tex]

[tex]Power = \frac{8*10^{6} J}{10}[/tex]

[tex]power = 8*10^{5} watt[/tex]

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proper question;

The kinetic energy of a car is 8 × 106 J as it travels along a horizontal road. How much power is required to stop the car in 10 s? (A) zero joules (B) 8  105 J (C) 8  107 J (D)8  104 J (E) 8  106 J​

The corpus callosum where fibers connect the brain's left and right hemispheres, thickens in adolescence, and this improves adolescents' ability to process information

The structure referred to in the statement is the corpus callosum. The corpus callosum is a broad band of nerve fibers that connects the left and right hemispheres of the brain. It plays a crucial role in facilitating communication and information exchange between the two hemispheres.

During adolescence, the corpus callosum undergoes a process called thickening, which refers to an increase in its size and structural integrity. This thickening is a result of ongoing myelination, a process where the nerve fibers are coated with a fatty substance called myelin. Myelination enhances the speed and efficiency of electrical impulses transmitted along the nerve fibers.

The thickening of the corpus callosum in adolescence has significant implications for cognitive functioning and information processing. It allows for enhanced coordination and integration of activities between the brain's hemispheres. The improved connectivity between the left and right hemispheres facilitates the transfer of information, enabling adolescents to utilize both sides of their brains more effectively.

As a result, adolescents experience improvements in various cognitive abilities, such as problem-solving, decision-making, and higher-order thinking skills. The enhanced communication between brain regions supports better coordination and integration of cognitive processes, leading to more efficient information processing.

In summary, the thickening of the corpus callosum during adolescence improves adolescents' ability to process information by enhancing communication and coordination between the brain's left and right hemispheres. This developmental change contributes to the cognitive advancements observed during this stage of life.

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(a) The difference between a transformer and an induction motor are:

(c) Four methods to start a three-phase induction motor are:

(d) Difference between a squirrel cage motor and a wound rotor motor:

Transformers use the power of electrical currents passing through coils to create a connection between the primary and secondary windings. Induction motors, on the other hand, use electromagnetic power to create a rotating magnetic force.

Note that a squirrel cage motor  has a rotor made up of conductive bars that are shorted at both ends.  The wound rotor motor has a part inside called the rotor that has three sets of wires connected to either outside resistors or slip rings.

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The collimation of the x-ray beam reduces scatter radiation and unnecessary exposure to the x-ray beam, and should be done to expose as large an area as possible (option d).

1. Collimation of the x-ray beam is an essential technique used in radiology to control the size and shape of the x-ray field.

2. One of the primary purposes of collimation is to reduce scatter radiation, which refers to the radiation that scatters in various directions within the patient's body. By limiting the size of the x-ray beam to the area of interest, collimation helps minimize scatter radiation, which can be harmful and reduce image quality.

3. Another crucial aspect of collimation is to minimize unnecessary exposure to the patient. By narrowing down the x-ray beam to the specific area being examined, areas outside the field of interest receive less radiation, reducing the patient's overall exposure.

4. It is important to note that collimation should be done to expose as large an area as possible while still maintaining adequate coverage of the region of interest. This ensures that the relevant structures are included in the x-ray image, providing valuable diagnostic information.

5. The statement that the collimation turns the beam at a 90-degree angle to x-ray patients that are standing (option C) is incorrect. Collimation does not involve changing the direction of the x-ray beam but rather controlling its size and shape.

6. The correct answer is option D: Both A and B. Collimation reduces scatter radiation and unnecessary exposure while also aiming to expose as large an area as possible, ensuring optimal imaging and patient safety.

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Answer:

- They need oxygen to burn fuel

- Aerodynamics

- Extreme temperatures

- Radiation

- Pressure issues

Explanation:

A airplane is a heavier-than-air aircraft kept aloft by the upward thrust exerted by the passing air on its fixed wings and driven by propellers, jet propulsion, etc.

Aeroplanes cannot travel in space for several reasons:

They need oxygen to burn fuel - Aeroplane engines rely on the oxygen in the atmosphere to burn fuel and generate thrust. In space, there is no atmosphere so there is no oxygen for the engines to work.

Aerodynamics - Aeroplane wings generate lift by interacting with the air. In space, there is no air so wings would be unable to generate any lift. Aeroplanes rely on aerodynamics to fly which does not work in space.

Extreme temperatures - In space, temperatures can range from -150 degrees Celsius to 150 degrees Celsius. Aeroplanes are designed to operate within a much narrower temperature range. The extreme cold and heat of space could damage aeroplane components.

Radiation - In space, there are high levels of radiation from the Sun and cosmic rays. Aeroplane bodies are not designed to shield against this type of radiation and it could damage electronics and affect aeroplane systems.

Pressure issues - Aeroplanes are designed to withstand air pressures at altitudes up to around 12 kilometers. In low-Earth orbit and beyond, the air pressure is essentially zero. This extreme change in pressure could cause structural damage to the aeroplane.

In summary, while aeroplanes are designed to fly through the Earth's atmosphere, they lack the key features needed to operate in the extreme environment of outer space like spaceships. Aeroplanes require things like oxygen, aerodynamics and being able to withstand changes in pressure - all of which do not exist or work the same way in space.

Explanation:

The wing is pushed up by the air under it. Large planes can only fly as high as about 7.5 miles. The air is too thin above that height. It would not hold the plane up.

Answer:

$20,900 decrease.

Explanation:

During the current year, Sue Shells, Inc.'s total liabilities decreased by $25,600 and stockholders' equity increased by $4,700. By what amount and in what direction did Sue's total assets change during the same time period? $20,900 decrease.

The resistance of a platinum resistance thermometer is 200 Ω at 0°C and 255.8 Ω at the melting point. Using the resistance-temperature relationship and calculations, the estimated melting point is approximately 19.93°C.

The resistance of a platinum resistance thermometer is 200 Ω when placed in a 0°C ice bath and 255.8 Ω when immersed in a crucible containing a melting substance. To determine the melting point of the substance, we need to calculate the temperature at which the resistance reaches 255.8 Ω.

First, we find the temperature coefficient of resistance (α) using the formula α = (R - R₀) / (R₀ * T), where R is the resistance at the melting point, R₀ is the resistance at 0°C, and T is the temperature at the melting point.

Substituting the given values, we have α = (255.8 - 200) / (200 * T₀), where T₀ is the known room temperature of 20°C.

Calculating α, we find α ≈ 0.014.

Next, we use the resistance-temperature relationship equation R = R₀(1 + αT) to solve for the melting point temperature (T). Substituting the known values, we have 255.8 = 200(1 + 0.014 * T).

Simplifying the equation, we find 1.279 = 1 + 0.014T.

Solving for T, we get T ≈ 19.93°C.

Therefore, based on the given data, the estimated melting point of the substance is approximately 19.93°C.

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Answer:

3400 K

Explanation:

use wiens displacement law

λ_max = 853 nm = 853 × 10^−9 m

T = b / λ_max

substiute

T = (2.898 × 10^−3 m·K) / (853 × 10^−9 m)

T = 2.898 × 10^−3 / 853 = 3.4 × 10^3 K

Answer:

Explanation:

The answer is **(c) 9.39 m/s** for the velocity of the bullet-block system after it's hit, and **(g) 209.39 m/s** for the bullet's velocity before it hit the box.

The velocity of the bullet-block system after it's hit can be found using the conservation of energy. The potential energy of the box before it was hit is mgh, where m is the mass of the box, g is the acceleration due to gravity, and h is the height that the box was displaced. After the bullet hits the box, the potential energy of the box is zero, but the kinetic energy of the bullet-block system is mv^2/2, where m is the total mass of the bullet-block system and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mgh = mv^2/2

```

Solving for v, we get:

```

v = sqrt(2mgh)

```

In this case, we have:

* m = 0.05 kg + 1.3 kg = 1.35 kg

* g = 9.8 m/s^2

* h = 4.5 m

So, the velocity of the bullet-block system after it's hit is:

```

v = sqrt(2 * 1.35 kg * 9.8 m/s^2 * 4.5 m) = 9.39 m/s

```

The velocity of the bullet before it hit the box can be found using the conservation of momentum. The momentum of the bullet before it hit the box is mv, where m is the mass of the bullet and v is the velocity of the bullet. After the bullet hits the box, the momentum of the bullet-block system is (m + M)v, where M is the mass of the box and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mv = (m + M)v

```

Solving for v, we get:

```

v = mv/(m + M)

```

In this case, we have:

* m = 0.05 kg

* M = 1.3 kg

* v = 9.39 m/s

So, the velocity of the bullet before it hit the box is:

```

v = 0.05 kg * 9.39 m/s / (0.05 kg + 1.3 kg) = 209.39 m/s

```

The velocity of the bullet-block system after the collision is approximately a) 6.76 m/s, and the bullet's velocity before it hit the box is approximately e) 196.76 m/s.

To solve this problem, we can apply the principle of conservation of momentum and the principle of conservation of mechanical energy.

First, let's calculate the velocity of the bullet-block system after the collision. We can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Let m1 be the mass of the bullet (0.05 kg) and m2 be the mass of the box (1.3 kg). Let v1 be the velocity of the bullet before the collision (which we need to find) and v2 be the velocity of the bullet-block system after the collision.

Using the conservation of momentum:

m1 * v1 = (m1 + m2) * v2

0.05 kg * v1 = (0.05 kg + 1.3 kg) * v2

0.05 kg * v1 = 1.35 kg * v2

Now, let's calculate the velocity of the bullet-block system (v2). Since the system goes up by a height of 4.5 m, we can use the principle of conservation of mechanical energy.

m1 * v1^2 = (m1 + m2) * v2^2 + m2 * g * h

0.05 kg * v1^2 = 1.35 kg * v2^2 + 1.3 kg * 9.8 m/s^2 * 4.5 m

Now, we can substitute the value of v2 from the momentum equation into the energy equation and solve for v1.

By solving these equations, we find that v1 is approximately 196.76 m/s.

Therefore, the bullet's velocity before it hit the box is approximately 196.76 m/s. (e) 196.76 m/s

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The displacement needed to get the ball into the hole on the first stroke is approximately 9.8731 m at an angle of 12.01° east of north

To calculate the displacement needed to get the ball into the hole on the first stroke, we need to consider the vector components of each stroke and combine them to find the resultant displacement.

First, we break down the second stroke of 3.0 m southeast into its north and east components. Since it's at a 45° angle, both components will have the same magnitude of 3.0 m.

North component = 3.0 m * cos(45°) = 3.0 m * 0.7071 ≈ 2.1213 m

East component = 3.0 m * sin(45°) = 3.0 m * 0.7071 ≈ 2.1213 m

Next, we break down the third stroke of 2.0 m southwest into its north and west components. Since it's at a 45° angle, both components will have the same magnitude of 2.0 m.

North component = 2.0 m * cos(45°) = 2.0 m * 0.7071 ≈ 1.4142 m

West component = 2.0 m * sin(45°) = 2.0 m * 0.7071 ≈ 1.4142 m

Now, we can sum up the north and east components from the first, second, and third strokes:

North displacement = 6.0 m + 2.1213 m + 1.4142 m = 9.5355 m

East displacement = 0 m + 2.1213 m - 0 m = 2.1213 m

Finally, we can calculate the magnitude and direction of the resultant displacement using the Pythagorean theorem and trigonometry:

Resultant displacement = √(North displacement^2 + East displacement^2) ≈ √(9.5355 m^2 + 2.1213 m^2) ≈ 9.8731 m

The direction can be found using the tangent function:

Direction = arctan(East displacement / North displacement) ≈ arctan(2.1213 m / 9.5355 m) ≈ 12.01°

Therefore, the displacement needed to get the ball into the hole on the first stroke is approximately 9.8731 m at an angle of 12.01° east of north.

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Answer:

(B) 29 m/s

Explanation:

The kinetic energy (KE) of an object can be calculated using the equation:

KE = (1/2) * m * v^2

where KE represents the kinetic energy, m represents the mass of the object, and v represents the velocity of the object.

Given that the mass of the truck is 1100 kg and the kinetic energy is 4.6 × 10^5 J, we can rearrange the equation to solve for the velocity (v):

KE = (1/2) * m * v^2

Solving for v:

v^2 = (2 * KE) / m

v^2 = (2 * 4.6 × 10^5 J) / 1100 kg

v^2 = 8.36 × 10^2 m^2/s^2

Taking the square root of both sides, we find:

v = √(8.36 × 10^2 m^2/s^2)

v ≈ 28.9 m/s

The speed of the truck is approximately 28.9 m/s.

Among the given options, the closest value to 28.9 m/s is (B) 29 m/s.

Answer:

B. Selective Pressures