Which scenario is an example of the transfer of thermal energy by
convection?
A. A person uses a hot iron to remove wrinkles from a shirt.
B. You warm your feet in front of a fireplace.
C. A cup heats up when you pour hot water into it.
D. Warm air in a hot air balloon rises and heats air at the top of the
balloon.

Answer:

F = 12000 N = 12 KN

Explanation:

First we need to find the acceleration of the bad boys by using second equation of motion:

s = Vi t + (1/2)at²

where,

s = distance covered = 5 m

Vi = Initial Velocity = 0 m/s

t = time taken = 0.5 s

a = acceleration = ?

Therefore,

5 m = (0 m/s)(0.5 s) + (1/2)a(0.5 s)²

a = 2(5 m)/(0.5 s)²

a = 40 m/s²

Now, we calculate the force applied by Newton's Second Law:

F = ma

where,

F = force = ?

m = mass of bad guys = 300 kg

Therefore,

F = (300 kg)(40 m/s²)

F = 12000 N = 12 KN

Answer:

the rocks have the same amount of thermal energy

Answer:

I personally think the answer is 5 because if you look at it its 5

\(\\ \sf\Rrightarrow F_{net}=F_1+F_2+F_3+F_4\)

\(\\ \sf\Rrightarrow F_{net}=20N+20N+10N-10N\)

\(\\ \sf\Rrightarrow F_{net}=40N\)

Answer:

\(force = mass \times acceleration \\ 100 = m \times 5 \\ m = \frac{100}{5} \\ m = 20 \: kg\)

Answer:

When an electromagnetic wave passes from space to matter, some part of the energy is absorbed by the matter and it increases its energy. The wave may reflect and some part may pass through the matter depending on the amount of energy they have. The amplitude of the wave decreases if some parts of it are reflected.

Electromagnetic radiation is one of the many ways that energy travels through space. The heat from a burning fire, the light from the sun, the X-rays used by your doctor, as well as the energy used to cook food in a microwave are all forms of electromagnetic radiation. While these forms of energy might seem quite different from one another, they are related in that they all exhibit wavelike properties.

Testable hypotheses are the theoretical problems that operational definitions have converted into.

Scientific observation is an empirical research designed to systematically address issues about the world. Through the use of operational definitions, theoretical questions are converted into. a provable theory. The measurement process for measuring external, observable behaviour is specified in Operational Definitions. The measurements that are obtained are utilised to define and quantify the construct. A measurement process (a collection of operations) is specified in an operational definition for the purpose of measuring an external, observable behaviour, and the definition and measurement of the hypothetical construct are then based on the results of the measurement procedure. No matter how solid your conceptual definition may be, you cannot measure anything without an operational definition.

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The maximum value of the average normal stress in the links connecting (a) points b and d is 101.56MPa and (b) points c and e is -21.7MPa.

Stress refers to the measure of an external force acting over the cross-sectional area of an object and is given as units of force per area: N/m2 (SI) or lb/in2 (US). The SI units are commonly referred to as Pascals (Pa). There are two types of stress experienced by normal stress and shear stress. When a force acts perpendicular to the surface of an object, it exerts a normal stress while when a force acts parallel to the surface of an object, it exerts a shear stress. In order to determine the average normal stress in the links provided, consider the bar ABC as a free body as attached. The moment of a force is the product of the force and the distance from the axis of rotation. Hence,

∑Mc = 0 where Mc is the summation of moment at point C

(0.04)FBD – (0.025 + 0.04)(20x10^3) = 0

(0.04)FBD – 1.3x10^3 = 0

FBD = 32.5x10^3 N (Link BD is in tension)

∑MB = 0 where MB is the summation of moment at point B

-(0.04)FCE – (0.025)(20x10^3) = 0

-(0.04)FCE – 0.5x10^3 = 0

FCE = -12.5x10^3 N (Link CE is in compression)

Net area of one link in tension = (0.008)(0.036-0.016) = 160x10^-6 m^2

Hence for two parallel links A = 320x10^-6 m^2

Net area of one link in compression = (0.008)(0.036) = 288x10^-6 m^2

Hence for two parallel links A = 576x10^-6 m^2

a) σBD = FBD/A = 32.5x10^3/320x10^-6 = 101.56x10^6 = 101.56MPa

b) σCE = FCE/A = -12.5x10^3/576x10^-6 = -21.7x10^6 = -21.7MPa

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Newton's laws apply to the following things in your list: galaxies, planets, rocks on Earth, trucks, satellites in space, rocks on Mars, and airplanes.

This is because Newton's laws are universal, governing the motion of objects and the forces that act upon them, regardless of their location or scale.

According to "Newton's first law of motion", if a body is in rest, it will stay in rest until or unless an external or an unbalanced force is applied on the body to make it move. A satellite has a forward thrust, which is offset by the gravity of the earth and keeps the satellite orbiting around its orbit not falling it into the earth. The momentum that the satellite gained from its launch combines with the gravity of the earth cause the satellite go into the orbit above earth.

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The resistance of the circuit is 24 ohms.

When an electric current flows through a conductor, it encounters resistance, which causes some of the electrical energy to be converted to heat. The amount of resistance depends on the properties of the conductor and the conditions of the circuit.

Ohm's law relates the voltage (V), current (I), and resistance (R) in a circuit, and is expressed as:

V = I x R

We can rearrange this equation to solve for resistance:

R = V / I

In this case, we are given the voltage (V) of 120 volts and the current (I) of 5 amps, so we can plug these values into the equation to calculate the resistance (R):

R = V / I

R = 120 V / 5 A

R = 24 ohms

Therefore, the resistance of the circuit is 24 ohms.

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If a kiloparsec (kpc) is equal to 1000 parsecs, light years is a kiloparsec equal to: 3, 260 light years.

The distance that light travels in a light-year is one year. At 186,000 miles (300,000 kilometres) per second and 5.88 trillion miles (9.46 trillion kilometres) per year, light travels through interstellar space.

Since a light-year is the maximum distance light can travel in one year and nothing can move faster than that, the answer appears to be 13.8 billion light-years.

Thus, 3,260 many light years is a kiloparsec.

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Answer:

In the given function b(t), the variable t represents the number of years after 2008. The   domain of this function is . The range more than 258.7 would not make sense. The graph of the function is always continuous

The given function b(t) shows the population of bobcats in northern Arizona since 2008. Therefore the variable t represents the number of years after 2008.

Since t represents the number of years after 2008, which is either positive or zero, therefore the domain of this function is .

The given function is a quadratic function and the coefficient of  is negative, so it is a downward parabola. The range above the y-coordinate of the vertex doesn't make any sense.

The vertex of parabola is defined by .

Vertex of the given function is (4.2,258.7).

Thus, the range more than 258.7 would not make sense for this function.

Since the given function is a polynomial function and polynomial functions are always continuous, therefore the graph of the function is always continuous

HOPE THIS HELPS <3

Frequency is inversely proportional to the time period. That is, f = 1/T.

This refers to the number of waves that passes a fixed place in a given amount of time.  Frequency is measured in hertz, this is equal to one event per second.

What is period, T ?

This refers to the time needed for one complete cycle of vibration to pass a given point. A period is measured in seconds.

Note:  AS the frequency of a wave increases, the period of the wave decreases. Frequency and Period are in reciprocal relationships.

Mathematically :

f = 1/T

T =Total time,Period

f =  Frequency

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Answer:

b Wind blowing West to East away from horse latitudes

Explanation: prevailing Westerlies are winds which blows away from the horse latitudes, this type of wind blow towards the poles between latitude 30 degrees to latitude 60degrees.

Prevailing Westerlies are known to be strongest in the winter months in the and weakest in the summer months

They are also very strong when the pressures are low over the poles and also very weak when the pressures are high over the poles.

The wall of the bulb of a thermometer is made thin beause allow the heat from sample being measured to reach the liquid quickly for fast response and quick reading (improves the quickness of response by the thermometer). ... — has a vacuum above the liquid column so there is no resistance to the expanding liquid.

Answer:

C

Explanation:

because that's how science works

Answer:

9 ohms

Explanation:

P = V²/R

R = v²/p = 6²/4 = 36/4 = 9 ohms.

The resistance is 9 Ohm.

To find the resistance, we need to know about the power in term of resistance and voltage.

Thus, we can conclude that the resistance is 9 Ohm.

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The gravitational potential energy will increase by 423.36 J

PE = mgh

PE₁ = 72 × 9.8 × 0

PE₁ = 0 J

PE = mgh

PE₂ = 72 × 9.8 × 0.6

PE₂= 423.36 J

Change in potential energy = PE₂ - PE₁

Change in potential energy = 423.36 - 0

Change in potential energy = 423.36 J

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An example of a project where an object with a high (k) value would be a benefit is one that has to do with the design of a rocket engine. Metals that have very high thermal conductivity are useful for building rocket engines. This is because of the amount of energy/heat that is required to be combusted by these parts to achieve a break from the earth's force field of gravity.

While engineering problems where having an object with a low (k) value would be a benefit is those that have to do with buildings. Buildings are made from concrete and brisk with have low thermal conductivity hence low

A k-value (also known as a k-factor or lambda value) is a measure of a material's thermal conductivity, or how quickly heat moves through it. It is a basic quality that is independent of material quantities.

Thermal conductivities of different building materials may be compared using k-values. This is often useful in determining the possibility of heat transmission between the inside and exterior of a structure.

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The total tension in the man's neck muscles while sliding down the water slide with his head raised and looking forward is 51.7 N.

When the man is sliding down the water slide, a centrifugal force acts on his body due to the circular motion. This force is directed outward and is equal to the mass of the man multiplied by the square of his velocity divided by the radius of the curve. Since the man is holding his head steady, the tension force in his neck muscles must counteract the centrifugal force.

To calculate the tension force, we can equate it to the centrifugal force. The centrifugal force is given by the formula F = (m * v^2) / r, where F is the centrifugal force, m is the mass of the man, v is his velocity, and r is the radius of the curve.

Given that the radius of the curve is 2.40 m and the speed of the man is 4.00 m/s, we can substitute these values into the formula to find the centrifugal force.

F = (m * (4.00 m/s)^2) / 2.40 m

F = (m * 16.00 m^2/s^2) / 2.40 m

F = (6.67 * m) N

Now we know that the centrifugal force is equal to 6.67 times the mass of the man.

Since the total tension force in the man's neck muscles is 54.5 N, we can equate this to the centrifugal force and solve for the mass of the man.

54.5 N = 6.67 * m

m = 54.5 N / 6.67 ≈ 8.17 kg

Now that we know the mass of the man, we can calculate the tension force in his neck muscles by substituting it into the centrifugal force formula.

F = (8.17 kg * 16.00 m^2/s^2) / 2.40 m

F ≈ 51.7 N

Therefore, the total tension force in the man's neck muscles while sliding down the water slide with his head raised and looking forward is approximately 51.7 N.

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Answer:

Answer:

it is decreased by 1/2

Explanation:

The velocity of the moving object is reduced by half decreasing speed is also decreasing the momentum of a certain object.

Answer:

-1/2

Explanation:

The DOH issues advisories from June to November due to increased health risks associated with the rainy season, disease outbreaks, and the need for preparedness and public education.

The Department of Health (DOH) issues public health advisories during the months of June to November for several reasons related to public health concerns specific to that time period. Here are some possible reasons:

1. Rainy Season: In many regions, June to November corresponds to the rainy season or monsoon season. During this time, there is an increase in rainfall, which can lead to various health risks such as waterborne diseases, flooding, and increased vector population (mosquitoes, rats, etc.). The DOH issues advisories to raise awareness and provide preventive measures against these risks.

2. Disease Outbreaks: Certain diseases are more prevalent or have a higher likelihood of outbreaks during specific months. For example, dengue fever is commonly associated with the rainy season due to the increased breeding of mosquitoes in stagnant water. Leptospirosis, which is transmitted through contaminated water, is also more common during this period. The DOH issues advisories to inform the public about these diseases and how to prevent them.

3. Preparedness and Response: The DOH aims to promote preparedness and timely response to public health emergencies. By issuing advisories in advance, they can inform the public about potential health risks, preventive measures, and necessary actions to take. This helps individuals and communities be better prepared and able to respond effectively to health threats.

4. Awareness and Education: Public health advisories serve as a means to educate and raise awareness about health-related issues. The DOH uses these opportunities to disseminate information on proper hygiene practices, sanitation, vaccination, and other preventive measures. By targeting specific months, they can address the specific health concerns associated with that time of the year.

It's important to note that the specific reasons for issuing advisories during these months may vary depending on the region and prevailing health risks. The DOH assesses local epidemiological data and collaborates with other agencies to determine the appropriate timing and content of the advisories.

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Answer:

Solid state of matter

Answer:

Solution

Given:

initial velocity (u)=0m/s (because an object

starts from the rest)

time (t)=3sec

diatance travelled(s)=14m(16-2=14)

acceleration (a)=?

Now,

According to the formula

s=ut+1/2at^2

or,14 =0×3+1/2×a×3^2

or, 14=0+a/2×9

or, 14=9a/2

or, 9a=28

or, a=28/9

or, a=3.1 m/s^2

Therefore, the acceleration of the object is 3.1m/s^2 ans.

The central region of the Milky Way, specifically the galactic bulge and halo, contains mostly old (Population II) stars and globular clusters. Population II stars are older, metal-poor stars that formed early in the universe's history, primarily consisting of hydrogen and helium.

These stars have lower masses and are typically found in globular clusters, which are densely packed, spherical collections of stars that orbit the galactic center.

The galactic bulge is the central, elongated region of the Milky Way, characterized by its high concentration of stars, gas, and dust. Population II stars are more abundant in this area due to their age and the early formation of the Milky Way. Similarly, globular clusters are often found in the bulge due to their strong gravitational pull towards the center of the galaxy.

In contrast, the galactic halo is the outermost region of the Milky Way, encompassing both the galactic disk and the bulge. While the halo is relatively sparse compared to the bulge, it is still home to a significant number of Population II stars and globular clusters. These ancient stars and clusters in the halo provide crucial insights into the formation and evolution of our galaxy.

In summary, the central region of the Milky Way, including the galactic bulge and halo, contains a majority of the old (Population II) stars and globular clusters, reflecting the early history and development of our galaxy.

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Answer:

The answer is C. anti social

Explanation:

Hope that helps!

Given

d: diameter

d = 41 cm

We need radius information so we will calculate it:

r: radius

r = d/2

r = 41/2

r = 20.5 cm

Rotating speed

w = 242 rpm

Procedure

At a distance r from the center of the rotation, a point on the object has a linear speed equal to the angular speed multiplied by the distance r. The units of linear speed are meters per second, m/s.

But before using the formula we need to have all the units in the same system. So we need to go from rpm to rad/s and from cm to m

Now we can calculate the linear velocity of the belt.

Answer

The linear velocity of the belt would be 5.2 m/s.