How can you explain why there are so many hours of darkness on December 21? How can you explain why there are so many hours of daylight on June 21?

Answer:

all I know is C

are there more questions? anyone?

-KARL IS STOOPID

Explanation:

1. The maximum displacement of the gram-negative bacterium during a 1-hour period, moving at a constant speed of 25 µm/s, would be 90 mm.

2. Assuming the Spring constant to be 1. The work done when the spring length is increased from 20 cm to 30 cm, is 0.5 J.

The bacterium's constant speed of 25 µm/s can be converted to mm/s by dividing by 1000, which gives us 0.025 mm/s. To find the maximum displacement, we multiply the speed by the time. In this case, the time is 1 hour, which is equivalent to 3600 seconds. Multiplying 0.025 mm/s by 3600 seconds gives us 90 mm. Therefore, the maximum displacement of the bacterium during a 1-hour period would be 90 mm.

The formula for calculating work done is given by \(W=\frac{1}{2} *k*(x_2^2-x_1^2)\), where W is the work done, k is the spring constant, \(x_2\) is the final length, and \(x_1\) is the initial length of the spring. In this case, the spring constant is 1, \(x_2\) is 30 cm (or 0.3 m), and \(x_1\) is 20 cm (or 0.2 m).

Substituting the values into the formula, we have W = (0.5)(1)(0.09- 0.04). Simplifying the equation, we get W = 0.025 J.

Therefore, the work done when the spring length is increased from 20 cm to 30 cm, with a spring constant of 1, is 0.025 J or 0.5 J when rounded to two decimal places.

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This statement is true. If the rock contained 6 grams of radium-226 at the closure temperature and 0.375 grams at the time of discovery, the remaining mass difference (6 - 0.375 = 5.625 grams) is attributed to the decay of radium-226 into radon-222. Therefore, the rock had 3.375 grams of radon-222 when it was discovered. Based on the given information, statement 5 is the only true statement.

To determine the statements that are true, let's consider the information provided.

The rock reached its closure temperature 6,400 years ago.

This statement is not true. The closure temperature is the temperature at which a mineral retains its daughter isotopes without any further loss or gain. However, the closure temperature is not directly related to the half-life of the parent isotope.

The rock reached its closure temperature 4,800 years ago.

This statement is not true. Similar to the previous statement, the closure temperature is not determined solely based on the half-life of the parent isotope.

The rock had 2.625 grams of radon-222 1,600 years ago.

This statement is not true. The amount of radon-222 present 1,600 years ago cannot be determined directly from the information provided.

When the rock was discovered, it had 5.625 grams of radon-222.

This statement is not true. The amount of radon-222 when the rock was discovered is given as 0.375 grams, not 5.625 grams.

When the rock was discovered, it had 3.375 grams of radon-222.

This statement is true. If the rock contained 6 grams of radium-226 at the closure temperature and 0.375 grams at the time of discovery, the remaining mass difference (6 - 0.375 = 5.625 grams) is attributed to the decay of radium-226 into radon-222. Therefore, the rock had 3.375 grams of radon-222 when it was discovered.

Based on the given information, statement 5 is the only true statement.

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The angle between the velocity vector and acceleration vector at any given point in the orbit would be 90 degrees, therefore the correct answer is option A.

The rate of change of the velocity with respect to time is known as the acceleration of the object. Generally, the unit of acceleration is considered as meter/seconds².

As given in the If a  satellite in orbit around Earth is in a uniform circular motion then we have to find the angle between the velocity vector and acceleration vector.

In a uniform circular motion, the angle between the velocity vector and acceleration vector is always π/2.

The angle between the velocity vector and acceleration vector at any given point in the orbit would be 90 degrees.

Thus, the correct answer is option A.

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(1) The magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is 8,000 m/s² east.

(2) The magnitude of the average force the Bat exerts on the ball while they are in contact is 1,120 N.

The given parameters:

The magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is calculated as follows:

\(a = \frac{v- u}{t} \\\\a = \frac{35 - (-45)}{0.01} \\\\a = \frac{80}{0.01} \\\\a = 8,000 \ m/s^2\)

The magnitude of the average force the Bat exerts on the ball while they are in contact is calculated as follows;

\(F = ma\\\\F = 0.14 \times 8,000\\\\F = 1,120 \ N\)

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Heat energy is carried by the motion of matter particles in convection. Heat is transferred during conduction by the vibration of the particles. Heat is directly transmitted by electromagnetic waves in radition.

The movement of internal energy occurs during Heat energy transfer. Conduction, convection, and radiation are the three methods of transferring heat energy. Radiation involves the movement of electromagnetic waves, convection involves the movement of warm particles, and conduction involves the direct contact of atoms. Heat is transferred by convection through a fluid, such as water or air. Heat is transferred together with the fluid (liquid or gas) as it flows from one place to another. Currents are the movements of hot masses of water or air. Heat is transferred by radiation using electromagnetic waves. Direct contact is how conduction transmits thermal energy. When two objects come into close proximity, thermal energy moves from the warmer object to the cooler object.

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Answer and

the answer is down below here

by movement in fluids

Explanation:

i did it on edge 2023

Answer:

the function of the blanket is to keep Kamins warm so he wouldn't get cold

Explanation:

the thick layer of the material acts as a body protector from the cold weather

Answer:

The function of the blanket is to provide warmth to Kamini, since the weather is cold they use a thick blanket.

Explanation:

There are approximately 134.7 slits per centimeter in the grating.

We will use the grating equation to find the number of slits per centimeter in the grating. Here's a step-by-step explanation:

1. Recall the grating equation: nλ = d sin(θ), where n is the order of the bright fringe, λ is the wavelength, d is the distance between slits, and θ is the angle of the bright fringe with respect to the normal.

2. In this problem, we are given the following information:
  - Wavelength (λ) = 650 nm
  - Angle with respect to the normal (θ) = 5°
  - The first bright fringe (n = 1, since we are excluding the central fringe)

3. Plug the given values into the grating equation:
  1 * (650 nm) = d * sin(5°)

4. Solve for the distance between slits (d):
  d = (650 nm) / sin(5°)
  d ≈ 7422.57 nm

5. Convert the distance between slits to the number of slits per centimeter (1 cm = 1,000,000 nm):
  Number of slits per centimeter = 1,000,000 nm/cm / 7422.57 nm/slit
  Number of slits per centimeter ≈ 134.7 slits/cm

So, there are approximately 134.7 slits per centimeter in the grating.

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The initial speed given to the rock was 28.18 m/s horizontally. The formula that is used is : he height, h = ut + (1/2)gt²

Given: Height, h = 39 m, Speed of sound, v = 343 m/s, Time, t = 2.98 s, Acceleration due to gravity, g = 9.8 m/s², Initial velocity of rock, u = ?

Formula: The height, h = ut + (1/2)gt² (Taking upward direction as positive)

We get, u = (h - (1/2)gt²)/t. The rock was thrown horizontally.

Therefore, there is no initial vertical velocity.

So, we have to consider only the horizontal velocity.

Hence, the formula of horizontal displacement, s is , s = v₀tAs

there is no acceleration in horizontal motion, v = v₀

Here, s = horizontal distance = ? v = 343 m/st

= 2.98 s

Substituting the given values, we get, h = (v₀/2)t

⇒ v₀

= 2h/t

g = 9.8 m/s²

Putting the values, we get, u = (39 - (1/2)(9.8)(2.98)²)/2.98

= 28.18 m/s

So, the initial speed given to the rock was 28.18 m/s horizontally.

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Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image​. Let v be the image distance. Using lens formula to find it :

\(\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\)

Put all the values,

\(\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm\)

So, the image distance from the lens is 24 cm.

Magnification,

\(m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2\)

The negative sign of magnification shows that the formed image is inverted.

The electric current flows through a wire, a magnetic field is generated around the wire. This magnetic field is circular and perpendicular to the direction of the current flow, meaning that it forms concentric circles around the wire.

The case of the wire in question, the magnetic field would also be circular and perpendicular to the direction of the current flow. However, since the wire is placed from the floor to the ceiling straight up and down, the magnetic field would also be vertically oriented. The strength of the magnetic field would depend on the magnitude of the current flowing through the wire. The greater the current, the stronger the magnetic field. Additionally, the distance from the wire would also affect the strength of the magnetic field. The field strength decreases as the distance from the wire increases. Overall, the magnetic field around the wire would be a circular, vertically oriented field that is proportional to the magnitude of the current flowing through the wire.

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Answer:

i dont know

Explanation:

The electric fields in the eight regions are as follows:

Region 1: E₁₂ = 0 (no charges)

Region 2: E₂₃ = -θ²/2ε₀

Region 3: E₃₄ = -θ³/2ε₀

Region 4: E₄₅ = -θ⁴/2ε₀

Region 5: E₅₆

To determine the electric field in the eight regions between the infinite parallel plates with uniform charge densities, we need to consider the superposition principle. According to the superposition principle, the total electric field at a point is the vector sum of the electric fields due to each individual charge.

Let's denote the charge densities of the plates as θ¹, θ², θ³, θ⁴, θ⁵, θ⁶, and θ⁷, where each θ represents the charge density of the corresponding plate. Since all the charges are negative, the electric fields due to each plate will point towards the plates.

In Region 1, between plates 1 and 2, there is no charge (θ¹ = 0). Therefore, the electric field in this region is zero.

In Region 2, between plates 2 and 3, the electric field is solely due to the charge density θ² of plate 2. Let's denote the electric field in this region as E₂₃. The electric field due to plate 2 is given by:

E₂₃ = -θ²/2ε₀

where ε₀ is the permittivity of free space.

Similarly, we can determine the electric field in each of the other regions:

In Region 3, between plates 3 and 4, the electric field is due to the charge density θ³ of plate 3. Let's denote the electric field in this region as E₃₄. The electric field due to plate 3 is given by:

E₃₄ = -θ³/2ε₀

In Region 4, between plates 4 and 5, the electric field is due to the charge density θ⁴ of plate 4. Let's denote the electric field in this region as E₄₅. The electric field due to plate 4 is given by:

E₄₅ = -θ⁴/2ε₀

Similarly, we can determine the electric fields for Regions 5, 6, and 7:

In Region 5, between plates 5 and 6, the electric field is due to the charge density θ⁵ of plate 5. Let's denote the electric field in this region as E₅₆. The electric field due to plate 5 is given by:

E₅₆ = -θ⁵/2ε₀

In Region 6, between plates 6 and 7, the electric field is due to the charge density θ⁶ of plate 6. Let's denote the electric field in this region as E₆₇. The electric field due to plate 6 is given by:

E₆₇ = -θ⁶/2ε₀

In Region 7, between plates 7 and infinity, the electric field is due to the charge density θ⁷ of plate 7. Let's denote the electric field in this region as E₇₈. The electric field due to plate 7 is given by:

E₇₈ = -θ⁷/2ε₀

Finally, in Region 8, beyond plate 7, there are no charges, so the electric field is zero.

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Answer:

Acceleration is (48-14)/40 m/s² = 34/40 m/s² = 0.85 m/s²

The acceleration acting on the train during this time of travel is 0.85m/s²

HOW TO CALCULATE ACCELERATION:

Where;

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The acceleration of John during the entire motion is 1.2 m/s².

The acceleration of John during the entire motion is calculated  by applying the following formula as shown below.

a = Δv / Δt

where;

a = ( v - u ) / ( t₂ - t₁ )

a = ( 4 m/s - - 2 m/s ) / ( 10 s - 5 s )

a = ( 6 m/s ) / ( 5 s )

a = 1.2 m/s²

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The complete question is below:

John walks away from his house at a constant speed of 2 m/s for 10 seconds , he then remembers he lost his key and immediately runs back to his house at a speed of 4 m/s in 5 seconds. find the average acceleration during the journey

Answer:

hhh

Explanation:

The speed of the block when it reaches the level portion of the surface will be the same as its initial speed. This is because there is no friction to slow it down, and no external forces acting on it horizontally to change its speed.

In the absence of friction and external horizontal forces, the block will continue to move with a constant velocity in a straight line. As it slides down the incline, gravity provides the only force acting on the block, which causes it to accelerate. However, when the block reaches the level portion of the surface, the incline's gravitational force component acting parallel to the surface becomes zero. Therefore, the block's velocity remains constant, meaning its speed doesn't change. Hence, the speed of the block when it reaches the level portion of the surface will be the same as its initial speed.

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Most Kenyan athletes do their practice in mountainous areas because it helps in developing their physical abilities and strengthening their lungs. It is believed that the altitude and tough terrain conditions make it possible to run more efficiently at lower altitudes. Training at high altitudes has several benefits.

First, the high-altitude environment has less oxygen, which means that athletes have to work harder to breathe and produce energy. This helps in developing their physical abilities and strengthening their lungs. As the body gets used to this environment, it is better equipped to handle oxygen and energy more efficiently, which can lead to improved performance. Second, training at high altitudes can lead to an increase in the number of red blood cells in the body. Red blood cells are responsible for carrying oxygen to the muscles, and an increase in their numbers can improve an athlete's endurance levels. Third, high-altitude training can also help in weight loss as it burns more calories than training at lower altitudes. This can help athletes maintain their weight and improve their fitness levels. Fourth, the tough terrain conditions in mountainous areas can also help athletes in developing their skills.

Running on uneven surfaces can help improve balance, agility, and coordination, which are essential for a runner. Training at high altitudes also has some drawbacks. The most common one is altitude sickness, which is caused by the lack of oxygen in the air. Altitude sickness can cause headaches, nausea, vomiting, and shortness of breath. It can also lead to dehydration, which can further worsen the symptoms. Additionally, high-altitude training can be expensive as athletes need to travel to mountainous areas to train. The cost of living and training in such areas can be high, which can be a barrier for some athletes.

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In the original Milgram study, what percentage of participants displayed complete compliance, administering shocks of 450 volts About 40% that they believed were lethal.

The Milgram experiment(s) on conformity to authoritative persons were a series of social psychology research carried out by Yale University psychologist Stanley Milgram. The willingness of 40 research participants, males from various occupations and educational levels, to follow orders from an authority figure that conflicted with their own moral values was examined.

The volunteers were led to believe that they were taking part in a different experiment in which they had to administer electric shocks to a "learner." If these electric shocks had been real, they would have increased over time to lethal levels.

Unexpectedly, the experiment revealed that a very high percentage of the lab rats would fully comply with the instructions, with every person reaching 300 volts and 65% reaching the full 450 volts.

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Answer:

hope this will help

please give brainliest if correct

Answer:

S = v0 t + 1/2 a t^2       a = 0 seems given in the problem

Vp  t = Vr t + 16

both travel for time t but the police car is going faster and travels 16 km farther

t (Vp - Vr) = 16

t = 16 km / (100 - 80) km/hr

t = 16 / 20 hr = 4/5 hr = 48 min

Answer:

m = 28.7[kg]

Explanation:

To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.

\(E_{k}=\frac{1}{2}*m*v^{2}\\\)

where:

Ek = kinetic energy = 1800 [J]

m = mass [kg]

v = 11.2 [m/s]

\(1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]\)

The maximum kinetic energy with which the same radiation ejects electrons from another metal with a work function of 2.19 eV is 1.31 eV.

When radiation of a certain wavelength falls on a metal surface, it can eject electrons from the surface if the energy of the radiation is greater than the work function of the metal. The work function is the minimum energy required to remove an electron from the metal surface. The maximum kinetic energy of the ejected electrons depends on the difference between the energy of the radiation and the work function of the metal. If the maximum kinetic energy of the ejected electrons is 0.60 eV for one metal with a work function of 2.90 eV, then the energy of the radiation can be calculated as 3.50 eV. Using this same radiation, the maximum kinetic energy of the ejected electrons for another metal with a work function of 2.19 eV can be calculated as 1.31 eV. This is because the difference between the energy of the radiation and the work function of the second metal is 3.50 eV - 2.19 eV = 1.31 eV.

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The moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel is 2.0 kgm².

The given parameters;

The moment of inertial of the wheel is calculated as follows;

\(I = I_{hoop} + 3I_{spoke}\\\\I = MR^2\ + \ 3(\frac{1}{3} mL^2)\\\\I = (4.3 \times 0.67^2) \ + \ (0.15 \times 0.67^2)\\\\I = 2.0 \ kgm^2\)

Thus, the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel is 2.0 kgm².

"Your question is not complete, it seems to be missing the following information";

find the moment of inertial of the wheel about an axis through its center and perpendicular to the plane of the wheel.

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Answer: Hello Miko here to help!

Explanation: The answer is c!