The transfer of energy by electromagnetic waves is called

Answers

Answer 1
The transfer of energy by electromagnetic waves is called electromagnetic radiation.

Electromagnetic waves can transfer energy through matter or across empty space.Nov 1, 2012
Answer 2
Final answer:

Radiation refers to the transfer of energy by electromagnetic waves, a process that can take place with no medium, such as the heat from the sun reaching Earth.

Explanation:

The transfer of energy by electromagnetic waves is known as radiation. This process does not require a medium; the energy is carried by photons in the electromagnetic waves. Examples of this process include the heat produced by the sun, which reaches the Earth via the transfer of radiant energy, and microwaves heating food through radiation.

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Related Questions

Radio Waves If your favorite FM radio station broadcasts at a frequency of 104.5 MHz, what is the wavelength of the station’s signal in meters? What is the energy of a photon of the station’s electromagnetic signal?

Answers

Answer:

The wavelength of the station’s signal is 2.9 meters

The energy of a photon of the station’s electromagnetic signal is [tex]6.9\times10^{-26}J [/tex]

Explanation:

Wavelength [tex] \lambda [/tex] is inversely proprtional to frequency (f) and directly proportional to velocity of the wave (v).

[tex]\lambda=\frac{v}{f} [/tex] (1)

But electromagnetic waves as radio signals travel at speed of light so using this on (1):

[tex]\lambda=\frac{c}{f}=\frac{3.0\times10^{8}}{104.5\times10^{6}}\approx2.9\,m [/tex]

Albert Einstein discovered that energy of electromagnetic waves was quantized in small discrete packages of energy called photons with energy:

[tex] E=hf=(6.6\times10^{-34})(104.5\times10^{6})\approx6.9\times10^{-26}J[/tex]

with h the Planck's constant.

Final answer:

The wavelength of a 104.5 MHz FM radio signal is approximately 2.87 meters, and the energy of a photon of this radio signal is approximately 6.92 * 10^-26 Joules.

Explanation:

The subject of this question is the relationship between the frequency, wavelength, and energy of radio waves, specifically those used in FM radio broadcasting.

To calculate the wavelength of the radio signal, one can employ the wave equation: velocity = frequency * wavelength. Since the velocity of electromagnetic waves, which include radio waves, is the speed of light (3 * 10^8 m/s), the wavelength can be obtained by rearranging the equation to: wavelength = velocity / frequency. Using your FM station's frequency of 104.5 MHz (or 104.5 * 10^6 Hz), the wavelength of the station's signal would be approximately 2.87 meters.

The energy of a photon from this radio signal could be found through the photon energy equation: E = h * f, where h is Planck's constant (6.62607004 × 10^-34 m^2 kg / s) and f is the frequency in Hz. Thus, the energy of a radio signal photon at 104.5 MHz would be approximately 6.92 * 10^-26 J.

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This is the body of law that codifies what a state defines as legal and/or illegal as well as the punishments for the violations of the laws.

Answers

Answer:

Criminal Law

Explanation:

Criminal Law is the body of law that codifies what a state defines as legal and/or illegal as well as the punishments for the violations of the laws. Examples of Criminal laws are murder, assault, theft, or drunken driving. Under criminal comes that procedure of prosecution of individual who  commit crime. It is different from the civil law.

Answer:

Criminal Law is the body of law that codifies what a state defines as legal and/or illegal as well as the punishments for the violations of the laws. Examples of Criminal laws are murder, assault, theft, or drunken driving. Under criminal comes that procedure of prosecution of individual who  commit crime. It is different from the civil law.

Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving East at 4.33 m/s. The two skaters entangle and move together across the ice. Determine the magnitude and direction of their post-collision velocity.

Answers

Answer:

V = 3.6385 m/s

θ = 47.46 degrees

Explanation:

the important data in the question is:

Skater 1:

[tex]M_1[/tex]= 39.6 kg

direction: south (axis y)

[tex]V_{1iy}[/tex] = 6.21 m/s

Skater 2:

[tex]M_2[/tex] = 52.1 kg

direction: east (axis x)

[tex]V_{2ix}[/tex] = 4.33 m/s

Now using the law of the conservation of linear momentum ( [tex]P_i = P_f[/tex] and knowing that the collision is inelastic we can do the next equations:

[tex]M_{1}V_{1ix}+M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex]  (eq. 1)

[tex]M_{1}V_{1iy}+M_2V_{2iy} = V_{sy}(M_1+M_2)[/tex]  (eq. 2)

Where [tex]V_{sx}[/tex] and [tex]V_{sy}[/tex] is the velocity of the sistem in x and y after the collision.

Note: the conservation of the linear momentum have to be make once by each axis.

Now, in the (eq. 1) the skater 1 don't have velocity in the axis x, so we can replace [tex]V_{1ix}[/tex] by 0 in the equation and get:

[tex]M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex]  (eq. 1)

also, in the (eq. 2) the skater 2 don't have velocity in the axis y, so we can replace [tex]V_{2iy}[/tex] by 0 in the equation and get:

[tex]M_{1}V_{1iy} = V_{sy}(M_1+M_2)[/tex]  (eq. 2)

Now, we just replace the data in both equations:

[tex](52.1)(4.33) = V_{sx}(39.6+52.1)[/tex]  (eq. 1)

[tex](39.6)(6.21) = V_{sy}(39.6+52.1)[/tex]  (eq. 2)

solving for [tex]V_{sx][/tex] and [tex]V_{sy}[/tex] we have:

[tex]V_{sx][/tex] = 2.46 m/s

[tex]V_{sy][/tex] = 2.681 m/s

using the pythagoras theorem we can find the magnitude of the velocity as:

V = [tex]\sqrt{2.46^2+2.681^2}[/tex]

V = 3.6385 m/s

For find the direction we just need to do this;

θ = [tex]tan^{-1}(\frac{2.681}{1.46})[/tex]

θ = 47.46 degrees

                     

 

Two wheels having the same radius and mass rotate at the same angular velocity ((Figure 1) ). One wheel is made with spokes so nearly all the mass is at the rim. The other is a solid disk.How do their rotational kinetic energies compare?A. The wheel with spokes has higher KE, but not twice as high.B. They are nearly the same.C. The solid wheel has higher KE, but not twice as high.D. The solid wheel has about twice the KE.E. The wheel with spokes has about twice the KE.

Answers

Answer:

E. The wheel with spokes has about twice the KE.

See explanation in: https://quizlet.com/100717504/physics-8-mc-flash-cards/

Answer:

Explanation:

We have to consider how the location of the mass affects the moment of inertia.

For a solid cylinder, I = mR²

For a hollow cylinder, I = 1/2mR²

Where

I ist the moment of inertia,

m is their masses,

R is the radius of rotation.

Since they have the same mass and radius, it can be seen that a hollow cylinder has twice the moment of inertia as a solid cylinder of the same mass and radius.

We know that the rotational kinetic energy is proportional to the moment of inertia. From;

Rotational KE = 1/2IW²

Where W is the angular speed.

so that at the same angular speed, the wheel with the spokes will have about double the kinetic energy as the solid cylinder. Take note that some of the mass is in the spokes so the moment of inertia is not exactly double.

Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of 1 m3 of water is 1000 kg.)

Answers

Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 [tex]m^{3}[/tex]

height through which it falls, h is 150 m

mass of 1 [tex]m^{3}[/tex] of water is 1000 kg

⇒mass of 1250 [tex]m^{3}[/tex] of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 [tex]\frac{m}{sec^{2} }[/tex]

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

        = 1250000×9.8×150 J

        = 1837500000 J

Electrical Energy = [tex]\frac{3}{4}[/tex](KE)

                            = [tex]\frac{3}{4}[/tex]×1837500000

                            = 1378125000 J per second

nder some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 6.0×105 km (comparable to our sun); its final radius is 16 km.If the original star rotated once in 35 days, find the angular speed of the neutron star.

Answers

Answer:

2.92 * 10³ rad/s

Explanation:

Given:

Initial Radius of Original Star (Ri) = 6.0 * 10^5 km

Final Radius of Neutron Star (Rf) = 16km

Angular Speed = 1 revolution in 35 days

We need to convert this to rad/s

To do that, we first convert to rad/day

i.e (1 rev/35 days) * (2π rad/ 1 rev)

We then convert the days to hour

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours)

Finally, we convert the hour to seconds (3600 seconds makes an hour)

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours) * (1 hour/ 3600 sec)

Angular Speed = 2π rad/ 3024000 secs

Angular Speed (wi) = 2.079 * 10^-6rad/s

From the question, we're asked to calculate the angular speed of the neutron star (wf)

Applying law of conservation of angular momentum to a system whose moment of Inertia changes, we have

Ii*wi = If * wf ----------------- Formula

Where Ii and If are the initial and final Inertia of the star

The relationship between Inertia and Radius of each object is I = 2/5MR²

So, Ii = 2/5(MRi²) and If = 2/5(MRf²)

Substitute the above in the formula quoted

We have 2/5(MRi²)wi = 2/5(MRf²)wf ---------------- Divide through by 2M/5

We are left with, Ri²wi = Rf²wf

Make wf the subject of the formula

wf = wi * (Ri/Rf)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km) * (6.0 * 10^5 km/16km)

wf = 2.92 * 10³ rad/s

A collapsible plastic bag contains a glucose solution. If the average gauge pressure in the vein is 1.33 x 103 Pa, what must be the minimum height h of the bag in order to infuse glucose into the vein? Assume the density of the solution is 1.02 x 103 kg/m3 .

Answers

Answer:

The height of the bag will be 0.133 m

Explanation:

We have given gauge pressure [tex]P=1.33\times 10^3Pa=1330Pa[/tex]

Density of solution [tex]\rho =1.02\times 10^3=1020kg/m^3[/tex]

We have to find the height of the bag

We know that gauge pressure is given by P=\rho gh

[tex]1330=1020\times 9.8\times h[/tex]

h=0.133m

So the height of the bag will be 0.133 m

Final answer:

The minimum height of the bag should be 0.245 meters to infuse glucose into the vein.

Explanation:

In order for the fluid to enter the vein, its pressure at entry must exceed the blood pressure in the vein. To find the height of the fluid, we need to convert the blood pressure in mm Hg to SI units. Since 1.0 mm Hg = 133 Pa, the blood pressure in the vein is 18 mm Hg above atmospheric pressure, which is equivalent to (18 mm Hg)(133 Pa/mm Hg) = 2386 Pa.

Now we can calculate the height of the fluid using the formula:

h = P/(ρg)

Where:

h is the height of the fluidP is the gauge pressure (2386 Pa)ρ is the density of the solution (1.02 x 10^3 kg/m^3)g is the acceleration due to gravity (9.8 m/s^2)

Substituting the given values into the formula, we find:

h = (2386 Pa)/((1.02 x 10^3 kg/m^3)(9.8 m/s^2)) = 0.245 m

The minimum height of the bag should be 0.245 meters in order to infuse glucose into the vein.

While a star is on the main sequence, its equilibrium is the result of the outward pressure of hot gas and the inward pressure of ________.

Answers

Answer:

Gravity

Explanation:

While a star is on the main sequence, its equilibrium is the result of the outward pressure of hot gas and the inward pressure of GRAVITY.

Gravity is a property of mass of objects. Pressure gravity is the resultant force felt by a particular mass object.

As gravity contact the star surface, density gradient is set up.

Pressure of gravity is a gravitational compression which increases the objects density, compresses the mass of objects and reduces the object's size.

A star might leave the sequence if the star run out of fuel for nuclear fusion in its core.

Blaise Pascal duplicated Torricelli's barometer using a red Bordeaux wine of density 965 kg/m3 as the working liquid (see figure below).
1. What was the height h of the wine column for normal atmospheric pressure?
2. Would you expect the vacuum above the column to be as good as that for mercury?
A. Yes
B. No

Answers

Answer:

a) h=10.701m

b) No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.

Explanation:

The Torricelli's experiment "was invented by the Italian scientist Evangelista Torricelli and the most important purpose of this experiment was to prove that the source of vacuum comes from atmospheric pressure"

Pressure is defined as "the force that is applied on any object in the direction perpendicular to the surface of the object in the unit area is known as the pressure. There are various types of pressure".

Part a

We have the density for the red Bordeaux wine given [tex]\rho=965\frac{kg}{m^3}[/tex], the atmospheric pressure on the Toriccelli's barometer is given by:

[tex]P_{atm}=\rho g h[/tex]

Solving for the height of wine in the column we have this:

[tex]h=\frac{P_{atm}}{\rho g}[/tex]

And replacing we have:

[tex]h=\frac{101300Pa}{965\frac{kg}{m^3} 9.81\frac{m}{s^2}}=10.701 m[/tex]

So the height of the red Bordeaux wine would be h=10.701m. A very high value on this case if we compare with the usual values for this variable.

Part b

No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.

Final answer:

The height of the wine column for normal atmospheric pressure can be calculated using the equation P = ρgh. The vacuum above the column in the wine barometer would not be as good as that for mercury. The correct answer is No (B).

Explanation:

The height h of the wine column for normal atmospheric pressure can be calculated using the equation P = ρgh, where P is the atmospheric pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height of the column. The atmospheric pressure can be taken as P = 101325 Pa and the acceleration due to gravity is approximately g = 9.8 m/s^2. Plugging in these values and the density of the wine, ρ = 965 kg/m^3, we can solve for h. Hence, h = P / (ρg).

To compare the vacuum above the column in the wine barometer with that of a mercury barometer, it is important to note that the height of the liquid column in a barometer is determined by the atmospheric pressure pushing down on the liquid. Since both the wine and mercury barometers are subjected to the same atmospheric pressure, the height of the columns will be the same if the densities of the liquids are the same. However, since the density of wine is less than that of mercury, the height of the wine column will be greater than that of the mercury column. Therefore, No (B) is the correct answer.

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A 75.5-kg person puts on a life jacket, jumps into the water, and floats. The jacket has a volume of 3.38 x 10-2 m3 and is completely submerged under the water. The volume of the person's body that is underwater is 6.42 x 10-2 m3. What is the density of the life jacket?

Answers

Answer:

Density of jacket will be [tex]680.4733kg/m^3[/tex]

Explanation:

We know that weight of water displaced= buoyant force=weight of man

Now volume of water displaced [tex]v=3.38\times 10^{-2}+6.42\times 10^{-2}=9.8\times 10^{-2}m^3[/tex]

Density of water [tex]d=100kg/m^3[/tex]

So weight of water displaced [tex]=9.8\times 10^{-2}\times 1000=98kg[/tex]

So weight of jacket = 98-75 = 23 kg

We have given volume of the jacket = [tex]3.38\times 10^{-2}m^3[/tex]

So density of jacket [tex]=\frac{mass}{volume}=\frac{23}{3.38\times 10^{-2}}=680.4733kg/m^3[/tex]

A potter's wheel (a solid, uniform disk) of mass 7.0 kg and .65 m radius spins about its central axis. A 2.1 kg lump of clay is dropped onto the wheel at a distance .41 m from the axis. Calculate the rotational inertia of the system.

a. 2.5 kg · m2
b. 1.8 kg · m2
c. 1.5 kg · m2
d. 0.40 kg · m2

Answers

Answer:

The rotational inertia of the system is 1.8 kg.m².

(b) is correct option.

Explanation:

Given that,

Mass of disk = 7.0 kg

Radius = 0.65 m

Mass of clay = 2.1 kg

Distance = 0.41 m

We need to calculate the rotational inertia of the system

Using formula of rotational inertia

[tex]I''=I+I'[/tex]

Where, I= the moment of inertia of a solid disk

I'=the moment of inertia of lump of clay

Put the value into the formula

[tex]I=\dfrac{MR^2}{2}+mr^2[/tex]

[tex]I=\dfrac{1}{2}\times7.0\times(0.65)^2+2.1\times0.41^2[/tex]

[tex]I=1.8\ kg.m^2[/tex]

Hence, The rotational inertia of the system is 1.8 kg.m².

The amount of electrons flowing through the wire is called what?



amperes (amps)


current


voltage


ohms

Answers

Answer:

Current

Explanation:

Current can refer to the flow of electrons through a conductor of some kind as well as the number of electrons flowing through the conductor.

Final answer:

The amount of electrons flowing through the wire is called current, which is measured in amperes or amps. Current is directly proportional to voltage and inversely proportional to resistance, as stated in Ohm's Law.

Explanation:

The amount of electrons flowing through the wire is called current. Electric current is measured in the unit known as ampere (A), which is defined as the flow of one coulomb of charge through an area in one second. According to Ohm's Law, the current I in a circuit is directly proportional to the voltage V applied across the circuit and inversely proportional to the resistance R of the circuit, which can be expressed by the equation I = V/R. The SI unit of current is ampere, whereas ohm is the SI unit for electrical resistance, which represents how strongly a material opposes the flow of electric current.

Would you expect the tensile strength of polychlorotrifluoroethylene to be greater, the same as, or less than that of a polytetrafluoroethylene specimen having the same molecular weight and degree of crystallinity?

Answers

Final answer:

The tensile strength of PCTFE is generally expected to be higher than that of PTFE, given the same molecular weight and degree of crystallinity, mainly due to stronger intermolecular forces from the chlorine atom in the PCTFE polymer chain.

Explanation:

The tensile strength of polychlorotrifluoroethylene (PCTFE) and polytetrafluoroethylene (PTFE) specimens having the same molecular weight and degree of crystallinity can vary. Although both materials are fluoropolymers and have similar structural characteristics, PCTFE tends to have a slightly higher tensile strength than PTFE. This is primarily because PCTFE has a chlorine atom in its polymer chain, which contributes to stronger intermolecular forces and therefore, higher tensile strength.

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rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 seconds. Determine the distance Rickey slides across the ground before touching the base.

Answers

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2[/tex]

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m[/tex]

The distance Rickey slides across the ground before touching the base is 15.825 m

Cosmic rays are particles traveling at extreme speeds through intergalactic space. Many are launched by exploding supernovae. Even the slowest cosmic rays travel at 43% of the speed of light. Calculate the gamma factor for cosmic rays traveling at these speeds.

Answers

Answer:

1.1

Explanation:

Cosmic rays are particles traveling at extreme speeds through intergalactic space. Many are launched by exploding supernovae. From the question, we know that slowest cosmic rays travel at 43% of the speed of light that is 43/100 multiply by the speed of light(speed of light= 3.0×10^8 m/s).

Gamma factor can be calculated using the formula below;

Gamma factor= 1/(1- v^2/c^2)^1/2 --------------------------------------------(1).

Gamma factor= 1/(1-3.0×10^8 metre per seconds ×43/100÷ 3.0×10^8 metre per seconds) ^1/2

Gamma factor= 1/(1-1.6641×10^16/3.0×10^2)^1/2

Gamma factor= 1/(1-1.16641×10^16/9×10^16)^1/2

Gamma factor= 1/(1-0.1296)^1/2

Gamma factor= 1/(0.87)^1/2

Gamma factor= 1/0.9327

Gamma factor= 1.072112535

Gamma factor= ~ 1.1

Cosmic rays are high-speed particles that travel through space close to the speed of light. The gamma factor for cosmic rays traveling at 43% of the speed of light is  1.17.

Cosmic rays are high-speed particles that travel through interstellar space at speeds close to the speed of light. When calculating the gamma factor for cosmic rays traveling at 43% of the speed of light, we can use the formula: γ = 1 / √(1 - v^2/c^2), where v is the velocity of the particles and c is the speed of light. Substituting the values, the gamma factor for cosmic rays traveling at 43% of the speed of light is approximately 1.17.

When an electron de-excites from the third quantum level to the second, and then to the ground state, two photons are emitted. The sum of the emitted frequencies equals the frequency of the single-photon that would be emitted if de-excitation were from the third to ____.

Answers

Answer:

ground state

Explanation:

  Lets take  

n=3 ,n=2 ,n=1 are the energy level.

Energy level n=1 is the ground energy level.

The energy from 3 to 1 = hν

The energy from 3 to 2 = hν₁

The energy from 2 to 1 = hν₂

We can say that

hν = hν₁ +  hν₂

If the electron were de-excitation from the third level to ground level then the sum of emitted frequency will be equal to the frequency of a single electron.

Therefore the answer is ground state.

Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.

Answers

Final answer:

Yes, the voltage across a capacitor in an RLC series circuit with AC can be greater than the source voltage, especially at resonance where the capacitive and inductive reactances cancel each other out and energy oscillates between the capacitor and inductor.

Explanation:

In an RLC series circuit with alternating current (AC), it is indeed possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source. This phenomenon occurs due to resonance in the circuit, where the capacitive and inductive reactances can cancel each other out at a certain frequency, known as the resonant frequency. When the circuit is at resonance, the voltages across the capacitor and inductor can be much greater than the source voltage because of the energy oscillating between the electric field of the capacitor and the magnetic field of the inductor.

According to the equation Vc = Q/C, where Q is the charge and C is the capacitance, we see that the voltage across the capacitor (Vc) is directly proportional to the amount of charge stored and inversely proportional to the capacitance. In an AC RLC circuit at resonance, the charge can oscillate at amplitudes resulting in voltages across the capacitor that exceed the source voltage.

The same can be true for the voltage across the inductor. If the circuit is driven at or near its resonant frequency, the inductive reactance and capacitive reactance can become equal in magnitude but opposite in phase, leading to a situation where the voltage across the inductor also exceeds the source voltage due to the energy stored in its magnetic field.

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Transform boundaries are classified under which type of fault?

Answers

Answer:

Strike-slip fault

Explanation:

Transform boundaries play the role of connecting the other plate boundary segments.

When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.

Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement  releases energy and takes place in a horizontal direction which is parallel to the force of compression.

Final answer:

Transform boundaries are classified under Transform Faults, where plates slide past each other causing stress and earthquakes, such as at the San Andreas Fault.

Explanation:

Transform boundaries are classified under the type of fault known as Transform Faults. These happen at plate boundaries where the plates are not moving away or toward each other, but instead sliding past one another. This movement causes a huge amount of stress, leading to earthquakes. The most recognized transform fault is the San Andreas Fault in California, where the Pacific Plate and the North American Plate are sliding past each other.

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Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 25.0 kg and that of the boat is 30.0 kg . (Take the package's direction of motion as positive.)

Answers

Answer:

V = -0.8 m/s

Explanation:

given,

mass of the child (m)= 25 Kg

mass of the boat(M) = 30 Kg

velocity of boat = ?

Assuming Boys throws package of mass(m₁) 6 Kg at the horizontal speed of       10 m/s

using conservation of momentum

(M + m + m₁) V = (M+ m)V + m₁ v

initial velocity V = 0 m/s

(M + m + m₁) x 0 = (M+ m)V + m₁ v

0 = (25+50)V + 6  x 1 0

75 V = -60

V = -0.8 m/s

negative direction shows that velocity in the direction opposite to the motion of package.

Final answer:

The velocity of the boat immediately after, assuming it was initially at rest, is 15.8 m/s.

Explanation:

To calculate the velocity of the boat, we can use vector addition. The boat's velocity relative to the water is perpendicular to the river's velocity. We can use the Pythagorean theorem to find the magnitude of the boat's velocity:

Vboat = sqrt((Vriver)2 + (Vboat)2)

Substituting the given values, we get:

Vboat = sqrt((5.0 m/s)2 + (15.0 m/s)2) = 15.8 m/s

Therefore, the velocity of the boat immediately after is 15.8 m/s.

Find the magnitude of the gravitational force this wire exerts on a point with mass m placed at the center of curvature of the semicircle.

Answers

Answer:

[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]

Explanation:

Assuming given,

Mass of wire be M

length of wire be L

small mass at center is = m

Radius of the wire be equal to = R = L/π

mass of small element of the wire

[tex]dM = \dfrac{M}{L}Rd\theta[/tex]

All the force are acting along y- direction

so, for force calculation

[tex]F = \int \dfrac{mdMG}{R^2} sin\theta [/tex]

[tex]F = \int_0^{\pi} \dfrac{m\dfrac{M}{L}RG}{R^2} sin\theta d\theta[/tex]

[tex]F = \int_0^{\pi} \dfrac{mMG}{L\dfrac{L}{\pi}} sin\theta d\theta[/tex]

[tex]F = \pi \dfrac{mMG}{L^2}}\int_0^{\pi}sin\theta d\theta[/tex]

[tex]F = \pi \dfrac{mMG}{L^2}}(-cos \theta)_0^{\pi}[/tex]

[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]

The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]

Given that :

mass of wire = M

length of wire = L

small mass at center = m

radius of wire = L / [tex]\pi[/tex]

Determine the magnitude of the gravitational force exerted by the wire on a point

First step : express the mass of small element of wire

dM = [tex]\frac{M}{L} Rd[/tex]∅

Since all forces act in the vertical direction the magnitude of the force exerted will be

F = [tex]\int\limits^\pi _o {\frac{m\frac{M}{L} RG}{R^2} } \, sin\beta d\beta[/tex]

[tex]F = \pi \frac{mMG}{L^2} \int\limits^\pi _0 {x} \, sin\beta d\beta[/tex]

Resolving equation above

Therefore F = [tex]2\pi \frac{mMG}{L^{2} }[/tex]

Hence we can conclude that The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]

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While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of 260 Hz; your friend begins to blow his horn as well, and you hear a beat frequency of 6.0Hz .

How fast is your friend approaching you?

Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

         [tex] v_s =7.74\ m/s[/tex]

Explanation:

given,

Speed of sound = 343 m/s

frequency of horn = 260 Hz

the friend is approaching, the frequency is increased by the Doppler Effect. The frequency is 266 Hz

using formula

         [tex]f' = \dfrac{v}{v-v_s}f_0[/tex]

         [tex]266= \dfrac{343}{343 - v_s}(260)[/tex]

         [tex]1.023= \dfrac{343}{343 - v_s}[/tex]

         [tex]343 - v_s = 335.26[/tex]

         [tex] v_s =7.74\ m/s[/tex]

the speed of friends approaching is equal to [tex] v_s =7.74\ m/s[/tex]

​ hot-air balloon is rising vertically. The angle of elevation from a point on level ground 122 feet from the balloon to a point directly under the passenger compartment changes from 17.9 degrees to 29.5 degrees. How​ far, to the nearest tenth of a​ foot, does the balloon rise during this​ period?

Answers

29.62 feet far, to the nearest tenth of a​ foot, does the balloon rise during this​ period

Explanation:

Hot-air balloon is rising vertically, the angle of elevation from a point on level ground = 122 feet (from the balloon)

The passenger compartment changes from = 17.9 degrees to 29.5 degree

              [tex]\tan \left(17.9^{\circ}\right)=\frac{a}{122 \mathrm{ft}}[/tex]

              [tex]a=122 \times \tan 17.9^{\circ}=122 \times 0.32299=39.4[/tex]

Similarly,

              [tex]\tan \left(29.5^{\circ}\right)=\frac{b}{122 f t}[/tex]

              [tex]b=122 \times \tan \left(29.5^{\circ}\right)=122 \times 0.56577=69.02[/tex]

The nearest tenth of a foot, the balloon rise during the period, as below

               69.02 – 39.4 = 29.62 ft.

A dolphin in seawater at a temperature of 25°C emits a sound wave directed toward the ocean floor 154 m below. How much time passes before it hears an echo? (The speed of sound in ocean water is 1533 m/s.)

Answers

Answer:

0.2s

Explanation:

SO for the dolphin to hear its echo, the sound wave must travel a distance twice as much as the displacement between the dolphin and the ocean floor. So d = 154 * 2 = 308 m

Since the speed of sound in ocean floor is v = 1533m/s we can find out the time by dividing the distance d by the speed of sound

t = d / v = 308 / 1533 = 0.2s

[tex](t = d / v )[/tex]The time passes before it hears an echo is 0.3secs

What is sound wave?

A sound wave  servers patterns of disturbance caused by the movement of energy traveling through a medium.

The speed of sound in ocean floor was given as [tex]( v = 1533m/s)[/tex]

To find the time, we can make use if the formula

[tex](t = d / v )[/tex]

Where t= time

v= velocity

d= distance

Then substitute ,we have

[tex]= 308 / 1533 = 0.2s[/tex]

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It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface?Part AA uniform solid cylinder.Part BA uniform sphere.Part CA thin-walled hollow sphere.Part DA hollow, cylinder with outer radius R and inner radius R/2.

Answers

Answer:

Part a)

[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]

B) uniform Sphere

[tex]f = \frac{2}{7}[/tex]

C) uniform hollow sphere

[tex]f = \frac{2}{5}[/tex]

D) uniform hollow cylinder with inner radius R/2 and outer radius R

[tex]f = \frac{5}{13}[/tex]

Explanation:

As we know that fraction of total energy as rotational energy is given as

[tex]f = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2}[/tex]

now we have

[tex] f = \frac{mk^2(\frac{v^2}{R^2})}{mv^2 + mk^2(\frac{v^2}{R^2})}[/tex]

[tex]f = \frac{\frac{k^2}{R^2}}{1 + \frac{k^2}{R^2}}[/tex]

now we have

A) uniform Solid cylinder

for cylinder we know that

[tex]\frac{k^2}{R^2} = 0.5[/tex]

[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]

B) uniform Sphere

for sphere we know that

[tex]\frac{k^2}{R^2} = \frac{2}{5}[/tex]

[tex]f = \frac{0.4}{1.4} = \frac{2}{7}[/tex]

C) uniform hollow sphere

for hollow sphere we know that

[tex]\frac{k^2}{R^2} = \frac{2}{3}[/tex]

[tex]f = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5}[/tex]

D) uniform hollow cylinder with inner radius R/2 and outer radius R

for annular cylinder we know that

[tex]\frac{k^2}{R^2} = \frac{5}{8}[/tex]

[tex]f = \frac{\frac{5}{8}}{\frac{13}{8}} = \frac{5}{13}[/tex]

The fraction of the total kinetic energy that is rotational for a uniform sphere is [tex]\frac{1}{3}[/tex]

How to calculate rotational kinetic energy.

Mathematically, the rotational kinetic energy of an object is given by this formula:

[tex]K.E_{rot}=\frac{1}{2} I\omega^2[/tex]  

Where:

I is the moment of inertia.[tex]\omega[/tex] angular velocity.

Since we know that half of the total kinetic energy for a hollow, cylindrical shell that is rolling without slipping on a horizontal surface is translational and the other half is rotational. Thus, this fraction is given by this mathematical expression:

[tex]K.E=\frac{\frac{k^2}{R^2} }{1+\frac{k^2}{R^2}}[/tex]

a. For a uniform sphere:

[tex]\frac{k^2}{R^2}=0.5[/tex]

Substituting, we have:

[tex]K.E=\frac{0.5 }{1+0.5}\\\\K.E=\frac{0.5 }{1.5}\\\\K.E =\frac{1}{3}[/tex]

b. For a thin-walled hollow sphere:

[tex]\frac{k^2}{R^2}=\frac{2}{5}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{2}{5} }{1+\frac{2}{5}}\\\\K.E=\frac{0.4 }{1.4}\\\\K.E =\frac{2}{7}[/tex]

c. For a uniform hollow sphere:

[tex]\frac{k^2}{R^2}=\frac{2}{3}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{2}{3} }{1+\frac{2}{3}}\\\\K.E=\frac{0.7 }{1.7}\\\\K.E =\frac{2}{5}[/tex]

d. For a hollow sphere with outer radius (R) and inner radius:

[tex]\frac{k^2}{R^2}=\frac{5}{8}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{5}{8} }{1+\frac{5}{8}}\\\\K.E=\frac{0.625 }{1.625}\\\\K.E =\frac{5}{13}[/tex]

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Three thermometers are in the same water bath. After thermal equilibrium is established, it is found that the Celsius thermometer reads 0 °C, the Fahrenheit thermometer reads 12°F, and the Kelvin thermometer reads 273 K. Which one of the following statements is the most reasonable conclusion?

a) the Kelvin thermometer is incorrect
b) the Celsius thermometer is incorrect
c) the Fahrenheit thermometer is incorrect
d) the three thermometers are all at different temperatures

Answers

Answer:

C) the Fahrenheit thermometer is incorrect

Explanation:

Since

1) K = °C + 273

2) °F = 9/5 °C + 32

for 0 °C

1) K = 0°C + 273 = 273 K

2) °F = 9/5 * 0°C + 32 = 32 °F

Thus the Kelvin thermometer measurement coincides with the Celsius measurement but not with the °F . On the other hand, if the Fahrenheit measurement is right, the Celsius thermometer and the Kelvin one should be wrong.

Therefore is more reasonable to assume that one thermometer failed (the one of Fahrenheit and both Kelvin and Celsius are right ) that 2 thermometers ( Celsius and Kelvin thermometers fail and the one of Fahrenheit is right)

For the three thermometers, we see that one thermometer failed and 2 are right,, Hence the Fahrenheit thermometer is incorrect

Option C is correct

What is ?

Generally, the equation for finding Kelvin is mathematically given as

K = °C + 273

Where

°F = 9/5 °C + 32

In conclusion, The Kelvin thermometer measurement relates with Celsius measurement but has no direct link with  Fahrenheit  F,  the Fahrenheit thermometer is incorrect

 

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A cylinder containing an ideal gas has a volume of 2.6 m3 and a pressure of 1.5 × 105 Pa at a temperature of 300 K. The cylinder is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K. The change in internal energy of the gas is +6.0 × 105 J. How much heat did the gas absorb?

Answers

Answer:[tex]13.5\times 10^{8}[/tex] joules

Explanation:

From the first law of thermodynamics,

Δ[tex]Q[/tex]=Δ[tex]U[/tex]+[tex]W[/tex]

Where [tex]Q[/tex] is the heat given to the gas,

[tex]U[/tex] is the internal energy of the gas,

[tex]W[/tex] is the workdone by the gas.

When pressure is constant,

[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]

[tex]V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}[/tex]

When pressure is constant,[tex]W=P[/tex]Δ[tex]V[/tex]

Where [tex]P[/tex] is pressure and [tex]V[/tex] is the volume of the gas.

Given [tex]P=1.5\times 10^{5}Pa[/tex]

Δ[tex]V=[/tex][tex]7.8-2.6=5.2m^{3}[/tex]

So,[tex]W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J[/tex]

Given that Δ[tex]U=6\times 10^{5}[/tex]

So,Δ[tex]Q=[/tex][tex]6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J[/tex]

Final answer:

By using the first law of thermodynamics, the ideal gas law, the given parameters and an additional calculation for the work done by the gas, we can calculate the total heat absorbed by the gas.

Explanation:

To calculate the heat absorbed by the gas, we use the first law of thermodynamics which states that the heat absorbed by a system is equal to the change in its internal energy plus the work done by the system on its surroundings, expressed as Q = ΔEint + W. The change in internal energy, ΔEint, is given as +6.0 × 10^5 J.

Since the pressure is constant, the work done by the gas, W, can be calculated using W = PΔV, where P is the pressure and ΔV is the change in volume. The change in volume can be determined using the ideal gas law before and after the change, PV = nRT, thus, ΔV = nR(ΔT)/P. Substituting the given pressure, temperature change from 300 K to 900 K, and ideal gas constant, we can find ΔV.

After plugging ΔV into the equation for work done, we then add this to the change in internal energy to find the heat absorbed.

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An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 4.8 m/s. A child on a bridge, 81 m above the river, sees the raft in the river below and attempts to drop a small stone onto the raft. The child releases the stone from rest. In order for the stone to hit the raft, what must be the horizontal distance between the raft and the bridge when the child releases the stone?

Answers

Answer:

19.5 m

Explanation:

t = Time taken

u = Initial velocity = 0 (Assumed thrown from rest)

s = Displacement = 81 m

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of linear motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 81=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{81\times 2}{9.81}}\\\Rightarrow t=4.06371\ s[/tex]

The time taken for the stone to reach the river is 4.06371 seconds

Distance = Speed×Time

[tex]Distance=4.8\times 4.06371=19.5\ m[/tex]

The horizontal distance between the raft and the bridge when the child releases the stone should be 19.5 m

Two platoons line up for a tug of war. If both platoons have the same number of participants on a team, what other factor is critical to either platoon's win?

A. age of men
b. height of men
c. total mass of the team
d. condition of the ground

Answers

Answer:

option (c)

Explanation:

In the game of tug of war, the Newton's third law is obeyed.

One team pulls the rope in one direction and the other team pulls the rope in another direction.

As the mass of one team is more, so it is harder to pull the rope by the another team.

So, it depends on the total mass of the team.

Option (c) is correct.

Answer:

The best answer would be C. Total mass of the team.

Explanation:

C would be the best answer because when looking in the formula of force then the 2 factors that are inputted are mass and acceleration but in this case acceleration would most likely not have much have an effect so that leaves the factor of mass.

Hope this helped!

A projectile is launched from the earth’s surface at initial speed v0 at angle θ0 with the horizontal. When the projectile is at its maximum height h, it has half the speed it had when it was at half its maximum height h 2 . At what angle was the projectile launched.

Answers

Answer:

θ₀ = 67.79°

Explanation:

Given info

we know that

Ymax = v₀y² / (2g)

v = v₀x  (when Y = Ymax)

when the projectile was at half its maximum height (Y')

v' = 2v = 2*v₀x

we can use the equations

(v')²= v'x² + v'y²   ⇒  (2*v₀x)² = (v₀x)² + v'y²   ⇒   v'y² = 3*(v₀x)²   (I)

if we know that

v'y² = v₀y² - 2*g*(y')  

y' = Ymax /2 = (v₀y² / (2g)) / 2 = v₀y² / (4g)

then

v'y² = v₀y² - 2*g*(v₀y² / (4g)) = v₀y² - (v₀y² / 2) = v₀y² / 2

v'y² = v₀y² / 2       (II)

we can say that (I) = (II)

3*(v₀x)² = v₀y² / 2   ⇒   v₀y = √6*v₀x

Finally we apply

tan θ₀ = v₀y / v₀x

⇒  tan θ₀ = √6*v₀x / v₀x = √6

⇒  θ₀ = tan⁻¹ (√6) = 67.79°

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flow on a planet where the acceleration of gravity varies with height so that g=g0−cz, where g0 and c are constants. Integrate "F=ma" along a streamline to obtain the equivalent of the Bernoulli equation for this flow.

Answers

Answer:

[tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]

Explanation:

first write the newtons second law:

F[tex]_{s}[/tex]=δma[tex]_{s}[/tex]

Applying bernoulli,s equation as follows:

∑[tex]δp+\frac{1}{2} ρδV^{2} +δγz=0\\[/tex]

Where, [tex]δp[/tex] is the pressure change across the streamline and [tex]V[/tex] is the fluid particle velocity

substitute [tex]ρg[/tex] for {tex]γ[/tex] and [tex]g_{0}-cz[/tex] for [tex]g[/tex]

[tex]dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0[/tex]

integrating the above equation using limits 1 and 2.

[tex]\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]

there the bernoulli equation for this flow is [tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]

note: [tex]ρ[/tex]=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

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