In the greenhouse effect, far infrared radiation is earth's surface and absorbed and reemitted by gases in the atmosphere. These gases have _____ in concentration over the past century.
a. Increased
b. Decreased
c. Remained constant
d. Varied unpredictably

Answer:    Answer: The object moves forward at 5 m/s, stops, and then changes velocity.

Explanation:

With the information given in the question we can graph the points (image attached).

As we can observe, in the first segment of the graph the velocity is increasing linearly (at a constant rate) and is 5 m/s, then in the second segment we can see the position of the object remains the same from second 2 to second 4, which means the object is stopped.

Finally, in the third and last segment, we can observe a change in velocity (at a negative constant rate, because is decreasing), which is decreasing until the object stops.

Explanation:

A man's mass and weight are measured on the Earth and the Moon. The mass of the man will remain same on the Earth and on the Moon.

Mass of an object is the measurement of the quantity of matter in an object. Since, the person would contain the same amount of matter in the body regardless of where he or she might be, the mass would be the same wherever it was measured on the Earth or the Moon.

The mass is a physical quantity which is associated with the amount of matter contained in an object. It is also related to the inertia of the object.

The weight of an object measures the pull on the object from the Earth or the gravitational force. An object feels heavy because of its weight. It is proportional to the gravitational force on the object. The greater is the weight of an object, the heavier the object will feel. Thus, the object will be heavier on the Earth and lighter on the Moon.

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The wavelength of an ultrasonic wave with a frequency of 1.0×10^5 Hz, given the speed of sound in air as 340 m/s, is approximately 3.4 mm.

The wavelength of a wave can be calculated using the formula λ = v/f, where λ represents the wavelength, v is the velocity or speed of the wave, and f is the frequency of the wave.

In this case, the frequency is given as 1.0×10^5 Hz, and the speed of sound in air is 340 m/s. By substituting these values into the formula, we can determine the wavelength.

λ = 340 m/s / (1.0×10^5 Hz) = 3.4×10^-3 m = 3.4 mm

Therefore, the wavelength of the ultrasonic wave emitted by bats, with a frequency of 1.0×10^5 Hz, is approximately 3.4 mm.

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Answer: 5. 9000 meters

6. 0.12626 miles

7. 26.78 meters

8. 1 mile per minute

9. 468 km per hour

Explanation:

5. 1000 m=1 km

9 km * 1000 m/1 km=

9 * 1000/1 = 9000 meters

6. 63360 inches = 1 mile

8000 inches * 1 mile/63360 inches=

8000/63360=0.12626 miles

7. 100 cm=1 m

2678 cm * 1 m/100 cm=

2678 * 1/100=

2678/100=26.78 meters

8. 60 minutes=1 hour

60 mph * 1 hour/60 minutes=

60 * 1/60=1 mile per minute

9. 1000 m=1 km

1 hour=60 minutes

1 minute=60 seconds

60 seconds * 60 minutes/1 hour=

60 * 60/1=3600 seconds

130 meters * 3600 seconds/1 second=

130 * 3600/1=468000 meters per hour

468000 m* 1 km/1000 m=

468000 * 1/1000=468 km per hour

The resultant velocity of the athlete is 19.7m/s at a bearing of 24.9⁰.

Step 1: Draw the vector diagram

The first step is to draw a vector diagram that depicts the athlete's velocity (18m/s due east) and the wind's velocity (8m/s at a bearing of 230⁰).

Step 2: Draw the resultant vector

Now, we draw the resultant vector from the tail of the first vector to the head of the second vector.

This gives us the resultant velocity of the athlete after being impacted by the wind.

Step 3: Calculate the magnitude and direction of the resultant vector

Using the triangle of vectors, we can calculate the magnitude and direction of the resultant vector.

The magnitude is the length of the vector, while the direction is the angle between the vector and the horizontal axis.

We can use trigonometry to calculate these values.

In this case, we have a right triangle, so we can use the Pythagorean theorem to calculate the magnitude of the resultant vector: \(R^{2} = (18m/s)^{2} + (8m/s)^{2} R^{2} = 324 + 64R^{2} = 388R = \sqrt{388R} = 19.7m/s\)

To calculate the direction of the resultant vector, we can use the inverse tangent function: Tanθ = Opposite/AdjacentTanθ = 8/18Tanθ = 0.444θ = tan⁻¹(0.444)θ = 24.9⁰

Therefore, the resultant velocity of the athlete is 19.7m/s at a bearing of 24.9⁰.

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The National Electric Code (NEC) identifies rigid metal conduit (RMC) as a wiring method, even though it is often referred to as galvanized rigid conduit (GRC).

If 11/2 RMC is threaded (screwed wrench-tight) together between boxes or enclosures, the maximum distance allowed between supports is 130 feet. This straight run of the conduit is almost 200 feet, so it would require additional support to ensure proper electrical performance and safety.

When determining the proper support spacing for a straight run of conduit, it is important to consider factors such as the weight of the conduit, the conductors inside, and the number of bends in the conduit. Too few supports can create sagging and tension, which can damage the insulation of the conductors and potentially cause a short circuit.

Too many supports can be costly and difficult to install. Therefore, it is important to understand the proper support spacing requirements and follow the NEC guidelines.

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The type of galaxy in which star formation is no longer occurring is an elliptical galaxy. Elliptical galaxies are made up of old stars and do not have much gas or dust, which are necessary for new star formation. In contrast, spiral galaxies have ongoing star formation as they have a lot of gas and dust in their arms.

However, it is important to note that there can be exceptions to this general rule and some elliptical galaxies may have some residual star formation occurring in certain regions.

In elliptical galaxies, star formation is no longer occurring. These galaxies consist mainly of older, low-mass stars and contain very little gas and dust, which are essential for new star formation. As a result, their stellar population is aging without being replaced by newly formed stars.

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The patient is experiencing acute anxiety due to surgery anticipation.

The patient's acute anxiety is a direct result of their anticipation of the surgery scheduled for the following day.

It is not uncommon for individuals to experience anxiety or fear in anticipation of medical procedures, especially if they have concerns about the surgery, anesthesia, potential pain, or the outcome.

The uncertainty and anticipation of the upcoming surgery can trigger intense emotional and psychological distress, leading to an acute anxiety attack.

In such cases, it is crucial for healthcare professionals to provide appropriate support and reassurance to the patient.

Open communication, empathy, and information about the procedure can help alleviate some of the patient's fears and anxiety.

Additionally, implementing relaxation techniques, such as deep breathing exercises or mindfulness techniques, may also assist in managing the acute anxiety symptoms.

Supporting the patient's emotional well-being and addressing their concerns can contribute to reducing their anxiety and facilitating a more positive surgical experience.

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Answer:

8.04 seconds

Explanation:

Assuming that the child starts from rest at the bottom of the hill and travels until she comes to a stop, we can use the following kinematic equation:

v_f^2 = v_i^2 + 2ad

where v_f is the final velocity (which is zero since the child comes to a stop), v_i is the initial velocity (which is the velocity at the bottom of the hill), a is the acceleration (-0.392 m/s²), and d is the distance traveled.

We can solve for d:

d = (v_f^2 - v_i^2) / (2a)

= (0 - v_i^2) / (2-0.392)

= v_i^2 / 0.784

Since the child is sliding along flat snow-covered ground, there is no change in elevation, so we can use the distance traveled from the bottom of the hill to the stopping point as the distance d.

To find the time it takes for the child to travel this distance, we can use the following kinematic equation:

d = v_it + 0.5a*t^2

where t is the time and all other variables are as previously defined.

Substituting the expression for d obtained above, we get:

v_i^2 / 0.784 = v_it + 0.5(-0.392)*t^2

Solving for t, we get:

t = (2 * v_i) / 0.392

We still need to find the value of v_i, the initial velocity of the child at the bottom of the hill. To do so, we can use conservation of energy. The child starts at rest at the top of the hill, so all the initial energy is potential energy. At the bottom of the hill, all the potential energy has been converted to kinetic energy. Assuming no energy is lost to friction, we can equate these two energies:

mgh = 0.5mv_i^2

where m is the mass of the child, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill.

Solving for v_i, we get:

v_i = √(2gh)

Substituting this expression for v_i into the expression for t obtained earlier, we get:

t = (2 * √(2gh)) / 0.392

Plugging in the values of g, h, and a, we get:

t = (2 * √(29.820)) / 0.392 = 8.04 seconds

Answer: P2 = (400 kPa x 1 L) / 3 L = 133.33 kPa

Explanation:

P1V1/T1 = P2V2/T2

P2 = (P1V1)/V2

plug in  values = P2 = (400 kPa x 1 L) / 3 L = 133.33 kPa

Although this computation is essentially based on a mathematical formalism or equation that integrates components, conversion factors are frequently employed to convert one type of measure into another.

The term "chemical equation" is frequently used in everyday chemistry. French chemist Jean Beguin established the first empirical formula in 1615. A chemical equation merely refers to a premise of a molecular reaction's process using symbols and figures.

1.5 days = 129600 second

5.2 ft      = 1.58496 meters

3600 s   = 1 hour

10.2 m    = 33 ft 5.57 in

305 g     = 0.305 Kg

180 pm   = 0.00000000018 m

73 kg      = 73000 g

1,366 s    = 22.76666667 min

86,000 m= 86 km

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The average density of the Earth is about 5.5 g/cm3. Since the crust has an average density of about 3.0 g/cm3, it suggests that the density of the Earth's interior is much higher than that of the crust.

According to the given information, the average density of the Earth is 5.5 g/cm3, and the crust has an average density of about 3.0 g/cm3. Since the density of the Earth's interior is not provided, we have to make an inference based on the given data. From this data, we can infer that the density of the Earth's interior is much higher than that of the crust because if the Earth's core were less dense than the crust.

 The average density of the Earth would be much less than 5.5 g/cm3.The density of the Earth's interior is higher than that of the crust. This suggests that the Earth's core must be made of denser materials such as iron, nickel, and other metals that are more dense than the minerals found in the crust.

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Answer:

.

Explanation:

26 = Terminal Velocity

27 = Force

34 = Momentum

33 = Laws of Conversation of Momentum (I think)

Answer:

180 [J].

Explanation:

1) the required work [W] can be calculated as difference of the energy: W=E₂-E₁, where E₁=mgh₁ - the energy before lifting, E₂=mgh₂ - the energy after lifting;

2) W=mgh₂-mgh₁, where m - mass; g=10 [N/kg], h - height;

3) then the required work [W]:

W=mg*(h₂-h₁)=30*6=180 [J].

The value of 'a' in the E = ak + c relationship for electrons in the conduction band of the tetravalent n-type semiconductor is approximately 5.9 x 10^-38 J - m^2.

We are given the equation E = ak + c, where E represents the energy of the electron, k is the wave vector, and c is a constant.

The cyclotron resonance for electrons occurs when the angular radiation frequency, ωe, is equal to the cyclotron frequency, ωc. In this case, we are given ωe = 2 x 10^11 rad/sec.

The cyclotron frequency is given by ωc = eB / m*, where e is the charge of the electron and m* is the effective mass of the electron.

The magnetic field, B, is given as 0.15 Wb/m^2.

We can rearrange the equation ωc = eB / m* to solve for m*:

m* = eB / ωc.

Plugging in the given values, we have m* = (1.6 x 10^-19 C) * (0.15 Wb/m^2) / (2 x 10^11 rad/sec).

Simplifying the expression, we find m* ≈ 1.2 x 10^-29 kg.

Now we can calculate the value of 'a' using the given cyclotron resonance condition.

Substituting E = ak + c and ωe = ωc into the equation, we have ak + c = ωe / eB * m*.

Rearranging the equation, we get a = (ωe / eB) * m*.

Plugging in the known values, we have a = (2 x 10^11 rad/sec) / [(1.6 x 10^-19 C) * (0.15 Wb/m^2)] * (1.2 x 10^-29 kg).

Simplifying the expression, we find a ≈ 5.9 x 10^-38 J - m^2.

Therefore, the value of 'a' in the E = ak + c relationship for electrons in the conduction band of the tetravalent n-type semiconductor is approximately 5.9 x 10^-38 J - m^2.

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8000 Joules is the amount of work required to haul up the equipment.

To calculate the work required to haul up the equipment, we need to consider two components: the work done against gravity and the work done against the weight of the rope.

The force due to gravity is given by the weight of the equipment, which is 100 N. The distance over which the force is applied is the height the equipment is being hauled, which is 40 meters. The work done against gravity can be calculated using the formula:

Work = Force × Distance

Work against gravity = 100 N × 40 m = 4000 N·m or 4000 J (Joules)

The weight of the rope can be calculated by multiplying the weight per meter (0.8 N/m) by the length of the rope (40 m):

Weight of the rope = 0.8 N/m × 40 m = 32 N

Since the rope is being hauled up, the work done against the weight of the rope is the same as the work done against gravity. Therefore, the work against the weight of the rope is also 4000 J.

The total work required to haul up the equipment is the sum of the work against gravity and the work against the weight of the rope:

Total work = Work against gravity + Work against rope weight

Total work = 4000 J + 4000 J

Total work = 8000 J

Therefore, it will take 8000 Joules of work to haul up the equipme

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A photographer focuses his camera on his subject. The subject then moves closer to the camera. To refocus, the lens should be moved further from the detector.

When the subject is moved closer to the camera, the camera's lens should be moved away from the detector. As the subject moves closer to the camera, the distance between the lens and the detector decreases, causing the camera to lose focus.

Therefore, in order to refocus the camera and capture a clear image of the subject, the lens must be moved further away from the detector, increasing the distance between the lens and the detector and reestablishing the proper focal length.

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This is due to the fact that occluded fronts combine two cold air masses, which causes one of the cold air masses to go downward.

When a warm air mass is sandwiched between two cold air masses, an occluded front occurs. In an occlusion, the warm front passes over the cold front, which dives beneath it.

In a front is obscured, the warm front is fully supplanted by the cold front, in which the warm air masses have completely disappeared. Furthermore, there are frequent shifts in the various weather producing circumstances because of the cold front's relatively low temperature.

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Answer:

Explanation:

Technology has both positive and negative impacts on the environment. Here are some examples of each:

Positive impacts:

Increased energy efficiency: Advances in technology have led to the development of more energy-efficient appliances, vehicles, and industrial processes, reducing energy consumption and greenhouse gas emissions.

Renewable energy: Technology has enabled the development of renewable energy sources such as wind and solar power, reducing our dependence on fossil fuels and reducing the impact of energy production on the environment.

Improved waste management: Technological innovations have improved waste management practices, making it easier to recycle and reduce waste.

Enhanced communication and transportation: Technology has improved communication and transportation, making it easier to access information and resources, reducing the need for travel, and minimizing the environmental impact of transportation.

Negative impacts:

Resource depletion: Technology often requires the extraction and use of natural resources such as minerals, oil, and gas, leading to resource depletion and environmental degradation.

Electronic waste: The increasing use of electronic devices has led to a growing problem of electronic waste, which can contain toxic materials and harm the environment if not disposed of properly.

Climate change: Some technologies, such as fossil fuel-based energy production and transportation, contribute significantly to climate change through the release of greenhouse gases into the atmosphere.

Habitat destruction: The development of technology and infrastructure often requires the destruction of natural habitats, leading to the loss of biodiversity and disruption of ecosystems.

Overall, technology has the potential to have both positive and negative impacts on the environment, and it is important to consider these impacts when developing and using technology in a way that is sustainable and equitable.

The disadvantages of refractive telescopes include chromatic aberration and longer size.

This was the telescope developed by Newton in 1668.It is the simplest type of telescope which has a tube with lenses on both ends. The light enters through the lens on the farther end of telescope. The light is refracted to a point near the lens on the near end. It is then then further magnified when seen through the lens on the eyepiece.

Even though it was more reliable, it was really heavy and had a larger aperture and longer body. Since high quality lenses were costly to make, cheaper lenses were used mostly. So there was chromatic aberration.

Therefore, the disadvantages of refractive telescopes include chromatic aberration and longer size.

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In wind, there is an energy conversion from potential energy to kinetic energy.

Potential energy is energy at a point. Kinetic energy refers to energy that is motion. Wind refers to air in motion. This implies that wind possesses kinetic energy.

Ultimately, we arrive at the fact that when we observe wind, there is an energy conversion from potential energy to kinetic energy.

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When a same-sized force is applied over a smaller area, the pressure increases.

Pressure is defined as the force acting per unit area (P = F/A). Here, F represents the force and A represents the area. When the area (A) decreases, and the force (F) remains constant, the overall pressure (P) increases according to the formula.

This concept can be easily understood by imagining a person standing on the ground wearing regular shoes versus standing on the ground wearing high heels. In both cases, the person's weight (force) remains the same, but the area of contact with the ground is much smaller with high heels. As a result, the pressure exerted on the ground by high heels is much higher than that of regular shoes.

Understanding the relationship between force, area, and pressure is crucial in various applications, such as hydraulic systems, pneumatic systems, and even in our daily lives. Increasing pressure by reducing the area can enhance the efficiency of certain processes while decreasing pressure by increasing the area can help distribute force more evenly and reduce potential damage or stress on surfaces.

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The work done during the process is  506.2 J.

The work done during the process is calculated by applying the following equation.

W = PΔV

W = P ( Vf - Vi )

where;

The final volume of the system is calculated as follows;

Vf = A x h

Vf = 7.4 x 10⁻³ m² x 7.2 x 10⁻² m

Vf = 5.33 x 10⁻⁴ m³

W = 9.5 x 10⁵ Pa x ( 5.33 x 10⁻⁴ m³ - 0 )

W = 506.2 J

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Answer:

See the answer below

Explanation:

Acceleration due to gravity (g) is the accelerational force experienced by a body as a result of the influence of the earth's gravitational force. It is a vector quantity, that is, it has both magnitude and direction.

The acceleration due to gravity is inversely proportional to the square of the radius of the earth. Hence, g at the equatorial region is slightly lesser (9.780 m/s2) than that of the polar regions (9.832 m/s2) of the earth because the earth bulges at the equator leading to a higher radius than at the poles.  

Given that T=ma and a= v^2/r, we can solve for the velocity, v = √(Tr/m)

To solve this problem, we need to use the concept of centripetal force, which is the force that keeps an object moving in a circular path. The centripetal force is given by the equation:

F = m * a

where F is the centripetal force, m is the mass of the object and a is the centripetal acceleration.

The centripetal acceleration is given by the equation:

a = v^2 / r

where v is the velocity of the object and r is the radius of the circular path.

The radius of the circular path in this case is the length of the string (12cm). The angle of 35° is not needed for solving the problem.

Now we can substitute the value of a into the equation for the centripetal force:

F = m * (v^2 / r)

We know the force is the tension of the string, T, and the mass of the object, m. So we can write

T = m * (v^2 / r)

Given that T=ma and a= v^2/r, we can solve for the velocity, v = √(Tr/m)

Without knowing the mass of the object and the tension of the string we can't find the velocity of the spinning ball.

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An electron in an s orbital can be identified by these four quantum numbers, which specify its location, orientation, and spin.

The four quantum numbers that can apply to electrons in s orbitals are:

1. Principal quantum number (n): The principal quantum number (n) represents the principal energy level of the electron and describes the size and energy of the electron’s orbital.

An s orbital is a subshell with a value of 0 for the second quantum number (l).

2. Azimuthal quantum number (l): For an s orbital, the azimuthal quantum number (l) has a value of 0. It identifies the type of orbital an electron occupies.

3. Magnetic quantum number (ml):

The magnetic quantum number (ml) specifies the orientation of the electron’s orbital around the nucleus.

For an s orbital, the value of ml is 0.4. Spin quantum number (ms): The spin quantum number (ms) indicates the spin of the electron.

The two possible values of the spin quantum number are +½ and -½.

The electron spin is denoted by an arrow pointing up or down, and it is represented by the letter m in the Schrödinger equation.

An electron in an s orbital can be identified by these four quantum numbers, which specify its location, orientation, and spin.

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Answer:

d

Explanation:

the answer of the above mentioned question is 2N to the left.

hope this helps!

Answer:

this is just and example of how to solve it

Explanation:

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