The transfer of heat by the movement of a fluid is called: Think about it: Imagine a pocket of air over the land (“land air”), and another pocket of air over the ocean (“ocean air”).

Which air pocket would you expect to heat up more during the day?

Why?

Answer: it would be 14.1 mm

Explanation: Sound is a wave. More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium. Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion.

A vibrating string produces a sound wave as illustrated in Figure 14.2, Figure 14.3, and Figure 14.4. As the string oscillates back and forth, part of the string’s energy goes into compressing and expanding the surrounding air. This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions. The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string. Some of the energy is lost in the form of thermal energy transferred to the air. You may recall from the chapter on waves that areas of compression and rarefaction in longitudinal waves (such as sound) are analogous to crests and troughs in transverse waves.

The frequency of sound waves is 144.1 m.

More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium.

Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion.

This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions. The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string.

Therefore, The frequency of sound waves is 144.1 m.

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Answer:

Weight of the box? 980 N

Value of the normal force? 630 N

Acceleration of the box? 7.5 m/s^2

Explanation:

I got the answer right

Substances like oil or graphite spray are examples of - lubricants.

Lubrication is the control of friction and a friction-reducing film between moving surfaces such as the given diagram. The lubricant used can be a fluid, solid, or plastic substance.

Thus, oil or graphite spray are examples of - lubricants.

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Answer:

The object takes approximately 1.180 seconds to complete one horizontal circle.  

Explanation:

From statement we know that the object is experimenting an Uniform Circular Motion, in which acceleration ([tex]a[/tex]), measured in meters per square second, is entirely centripetal and is expressed as:

[tex]a = \frac{4\pi^{2}\cdot R}{T^{2}}[/tex] (1)

Where:

[tex]T[/tex] - Period of rotation, measured in seconds.

[tex]R[/tex] - Radius of rotation, measured in meters.

If we know that [tex]a = 26.36\,\frac{m}{s^{2}}[/tex] and [tex]R = 0.93\,m[/tex], then the time taken by the object to complete one revolution is:

[tex]T^{2} = \frac{4\pi^{2}\cdot R}{a}[/tex]

[tex]T = 2\pi\cdot \sqrt{\frac{R}{a} }[/tex]

[tex]T = 2\pi\cdot \sqrt{\frac{0.93\,m}{26.36\,\frac{m}{s^{2}} } }[/tex]

[tex]T \approx 1.180\,s[/tex]

The object takes approximately 1.180 seconds to complete one horizontal circle.  

Answer:

The helium molecules move closer together when it's cold causing the balloon to deflate

Explanation:

The possible reason behind it is: the temperature of outside was low, so the Helium gas inside the balloon compressed and  it was partially deflated at the car. When she returned to the house, temperature of the gas of the balloon rises again and  it was fully inflated again.

The experimental gas law known as Charles' law, commonly referred to as the law of volumes, illustrates how gases tend to expand when heated. Charles's law can now be expressed as follows:

The Kelvin temperature and the volume will be directly proportional when the pressure on a sample of a dry gas is kept constant.

This direct proportional relationship can be expressed as follows:

V∝T.

Hence, as the temperature is low outside the house, the balloon was partially deflated at the car and the temperature rises again when it was inside the house, so  it was fully inflated again inside the house.

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The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

[tex]P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\[/tex]

Where:

Solving for v0b:

[tex]v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}[/tex]

The original velocity of the ball is 5.71 m/s.

The principle of conservation of momentum: In a closed system, The total momentum before collision is equal to total momentum after collision.

From the principle of conservation of momentum,

MU+mu = MV+mv.................... Equation 1

Where M = mass of the bowling ball, m = mass of the pin, U = initial velocity of the bowling ball, u = initial velocity of the pin, V = final velocity of the bowling ball, v = final velocity of the pin.

From the question,

Given: M = 7 kg, m = 2 kg, u = 0 m/s (at rest), v = 6.0 m/s, V = 4 m/s.

Substitute these values into equation 1 and solve for U

7(U)+2(0) = 7(4)+2(6)

7U = 28+12

7U = 40

U = 40/7

U = 5.71 m/s.

Hence, The original velocity of the ball is 5.71 m/s.

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Answer:

I think white is the colour of star A.

Answer:

yellow

Explanation:

Answer:

Explanation:

What is flow rate?

Flow rate is the volume of fluid per unit time flowing past a point through the area  

the formula is Q=Av

where A= area

         v= flow velocity

step one:

given data

radius r1= 3cm= 0.03m

radius r2=2cm= 0.02m

velocity v1=?

velocity v2=5m/s

The area A1

[tex]Area= \pi r^2\\\\Area- A1= 3.142*0.03^2\\\\A1= 0.0028m^2[/tex]

The area A2

[tex]Area= \pi r^2\\\\Area- A2= 3.142*0.02^2\\\\A1= 0.0013m^2[/tex]

Step two:

For the continuous flow

Q1=Q2

A1v1=A2v2

0.0028*v1=0.0013*5

v1=0.0065/0.0028

v1=2.32m/s

The bloods velocity in the 3 cm radius unconstricted region is 2.32m/s

A maritime polar air mass has relatively higher moisture and is relatively cooler than the tropical air mass.

Maritime polar (mP) is cold but moist due to its origination over the oceans. The desert region air masses (hot and dry) are designated by ‘cT’ for ‘continental tropical’. As these air masses move around the earth they can begin to acquire additional attributes.

The continental Tropical (cT) air mass originates in arid or desert regions in the middle or lower latitudes, principally during the summer season. It is strongly heated in general, but its moisture content is very low.

Continental tropical air masses are extremely hot and dry. Arctic, Antarctic, and polar air masses are cold. The qualities of arctic air are developed over ice and snow-covered ground. Arctic air is deeply cold, colder than polar air masses

Maritime Polar (mP) air masses develop over the polar areas of both the Northern and the Southern hemispheres. They generally contain considerably more moisture than the cP air masses. As they move inland in middle and high latitudes, heavy precipitation may occur.

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Answer:

A planet's orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun's gravitational pull on it, and the faster the planet moves.

Explanation: hope that helps :)

Answer:

The answer is "Slowing down ".

Explanation:

please find the complete question in the attached file.

In this question, if the block B weight were accounted for by kinetic the friction of frame A, because the blocks pushed at a consistent speed throughout the beginning.

Afterward, on block A, the resistance intensity rises, which allows frames to also be negative, which is defined in the graph, that's why the answer Slowing down is correct.

Answer:

Explanation:

Using the equation of motion

V² = u²+2g∆H

v is the final velocity

u is the initial velocity = 0m/s

g = 9.8m/s²

∆H = 12.7-1.5 = 11.2m

Substitute

v² =0²+2(9.8)(11.2)

v² = 219.52

v = √219.52

v = 14.82m/s

Hence the final.speed of the ball is 14.82m/s

Answer:

D

Explanation:

Magnetic domains are always present in ferromagnetic materials due to the way the atoms bond to form the material. However, when a ferromagnetic material is in the unmagnetized condition, the magnetic domains are randomly oriented so that the magnetic field strength in the piece of material is zero.

Answer:

D. magnetic fields of atoms being aligned with each other

Explanation:

Answer: Use the x/t = v formula!!! x is distance, t is time, and v is velocity. you just need to do 40 / 5 = 8 and 100 / 15 = 6.66

Answer:

340m/s

Explanation:

Given parameters:

Wavelength  = 2m

Frequency   = 170Hz

Unknown:

Velocity of the wave  = ?

Solution:

The velocity of the wave is the product of frequency and wavelength of the wave;

          V  = F λ

V is the velocity

 F is the frequency

λ is the wavelength

  Insert the parameters and solve;

          V = 2  x 170 = 340m/s

The velocity of the wave with the given wavelength and frequency is 340m/s.

Given the data in the question;

Velocity; [tex]v =\ ?[/tex]

To determine the velocity of the wave, we use the expression for the relations between wavelength, frequency and speed.

[tex]\lambda = \frac{v}{f}[/tex]

Where [tex]\lambda[/tex] is wavelength, f is frequency and v is speed.

We substitute our given values into the equation

[tex]2m = \frac{v}{170s^{-1}}\\\\v = 2m * 170s^{-1}\\\\v = 340m/s[/tex]

Therefore, the velocity of the wave with the given wavelength and frequency is 340m/s.

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Answer:

increase distance

Explanation:

Inclined planes make it easier to move objects to a higher elevation. The sloping surface of the inclined plane supports part of the weight of the object as it moves up the slope. As a result, it takes less force to move the object uphill.

Answer:

Hello your answer would be increase of distance

Explanation:

I know that because than you would have to go upward with the object.

Answer:

His initial velocity (speed) must be 64.16 m/s.

Explanation:

Horizontal Motion

When an object is horizontally launched at a speed v from a height h, the maximum horizontal distance (also called Range) traveled by the object can be calculated as follows:

[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]

Our daredevil is attempting to clear a river d=420 m wide with his motorcycle from a cliff h=210 m high. We need to find the speed needed to safely reach the opposite edge of the river.

From the above equation, we solve for v:

Speed:

[tex]\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}[/tex]

[tex]\displaystyle v=420\cdot\sqrt{\frac {9.8}{2\cdot 210}}[/tex]

[tex]\displaystyle v=420\cdot\sqrt{\frac {9.8}{420}}[/tex]

v = 64.16 m/s

His initial velocity (speed) must be 64.16 m/s.

Answer:

Explanation:

Positive values for position indicate that the object is in front of the starting point and negative values tell us that the object is behind the starting point. (time = 9.5, position = 0) the object is at the starting point.

Answer:

Positive values for position indicate that the object is in front of the starting point and negative values tell us that the object is behind the starting point. (time = 9.5, position = 0) the object is at the starting point.

Explanation:

Answer:

Option D (On the...............dominate) would be the right approach.

Explanation:

The Gravitational constant (G) will be:

= [tex]6.67\times 10^{-11}[/tex]

The Coulomb's law constant (K) will be:

= [tex]9\times 10^9[/tex]

⇒ [tex]\frac{f_e}{f_g}\sim 10^{42}[/tex]

The latter three choices aren't connected to either the situation mentioned in the clarification segment elsewhere here.

The statement that provides evidence that there must be at least two types of electrical charge, but that there is only one type of mass is that on the scale of the solar system, electric forces cancel out and gravitational forces dominate.

The examples that provides conclusive evidence of is option d because the Gravitational constant is (G)=6.67*10-11 while the Coulomb's law constant(K)=9*109.

The interaction between charged objects is regarded as a non-contact force that acts over some distance of separation.

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Answer:

um were is it

Explanation:

Answer:

i think its b

Explanation:

Answer:

The potential is  [tex]V_A = 9600 \ V[/tex]

Explanation:

From the question we are told that  

   The  magnitude of the charge is  [tex]q_1 = 4 \mu C = 4*10^{-6} \ C[/tex]

   The position of the charge is  [tex]x = + 3.0 \ m[/tex]

   The magnitude of the second charge is  [tex]q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C[/tex]

   The position is  [tex]y_1 = - 1.0 \ m[/tex]

     The position of point A is  [tex]y_2 = + 4.0 \ m[/tex]  

Generally the electric potential  at A due to the first charge is mathematically represented as

         [tex]V_a = \frac{k * q_1 }{r_1 }[/tex]

Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]

        [tex]r_1[/tex]  is the distance between first charge and a which is mathematically represented as

         [tex]r_1 = \sqrt{x^2 + y_2 ^2 }[/tex]

=>      [tex]r_1 = \sqrt{3^2 + 4 ^2 }[/tex]    

=>      [tex]r_1 = 5 \ m[/tex]    

So

           [tex]V_a = \frac{9*10^9 * 4*10^{-6} }{5 }[/tex]

      [tex]V_a = 7200 \ V[/tex]

Generally the electric potential  at A due to the second charge is mathematically represented as

         [tex]V_b = \frac{k * q_2 }{r_2 }[/tex]

Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]

        [tex]r_2[/tex]  is the distance between second charge and a which is mathematically represented as

         [tex]r_2 = y_2 - y[/tex]

=>      [tex]r _2 = 4.0 - (-1.0)[/tex]    

=>      [tex]r = 5 \ m[/tex]    

So

           [tex]V_a = \frac{9*10^9 * -2*10^{-6} }{5 }[/tex]

      [tex]V_a = -3600 \ V[/tex]

So the net potential difference at point A  due to the charges is mathematically represented as

       [tex]V_n = V_a + V_b[/tex]

=>     [tex]V_n = 7200 - 3600[/tex]

=>     [tex]V_n = 3600 V[/tex]

Generally the net potential difference at the origin due to both charges is mathematically represented as

     [tex]V_N = V_c + V_d[/tex]

Here  

      [tex]V_c = \frac{k * q_1 }{x}[/tex]

=>   [tex]V_c = \frac{9*10^9 * 4*10^{-6} }{3}[/tex]

=>   [tex]V_c = 12000 V[/tex]

and

              [tex]V_d= \frac{k * q_2 }{y}[/tex]

=>   [tex]V_c = \frac{9*10^9 * -2*10^{-6} }{1}[/tex]

=>   [tex]V_c =- 18000 V[/tex]

Generally the net potential difference at the origin is  

       [tex]V_N = 12000 - 18000[/tex]

=>     [tex]V_N = -6000[/tex]

Generally the potential difference at A relative to zero at the origin is mathematically evaluated as

         [tex]V_A = V_n - V_N[/tex]

=>      [tex]V_A = 3600 - (-6000)[/tex]

=>      [tex]V_A = 9600 \ V[/tex]

The wave which has the longest frequency if, listed from smallest to largest are gamma ray, x-rays, ultraviolet, visible, infrared, microwave, and radio is radio waves and gamma rays have the highest frequency.

The oscillations between an electric field and a magnetic field produce waves known as electromagnetic waves or EM waves. In other words, magnetic and electric fields oscillate to form electromagnetic (EM) waves.

The term "electromagnetic spectrum" (EMS) refers to the entire range of electromagnetic radiation that is now understood. Wavelengths rise from roughly 10⁻¹⁸ m to 100 km, and frequencies fall from 3 × 10²⁶ Hz to 3 × 10³ Hz in this range.

The following list of the seven electromagnetic waves is organized by decreasing wavelength and increasing frequency and energy: X-rays, gamma rays, UV rays, visible light, radio waves, microwaves, and infrared waves.

Thus, radio waves have the longest frequency and gamma rays have the highest frequency.

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Linear acceleration and angular acceleration are directly proportional. The greater the linear acceleration, the larger the angular acceleration.

For example, the greater the angular acceleration of a bike's wheels, the greater the acceleration of the bike.