~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)


a.) Determine the linear regression

b.) At this given angular velocity, what is the rotational kinetic energy?

Answers

Answer 1

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J


Related Questions

g a horizontal wheel of radius is rotating about a vertical axis. What is the magnitude of the resultant acceleration of a bug that is hanging tightl on the rim of the wheel

Answers

Answer:

   a = w² r

Explanation:

In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.

           a = v² / r

angular and linear variables are related

           v = w r

we substitute

          a = w² r

where r is the radius of the wheel

Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.

Diagram A
Diagram B
Diagram C
Diagram D

Answers

Answer:Diagram A

Explanation:

Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.

what is time taken by radio wave to go and return back from communication satellite to earth??​

Answers

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

hope it help

In a certain cyclotron a proton moves in a circle of radius 0.530 m. The magnitude of the magnetic field is 1.30 T. (a) What is the oscillator frequency

Answers

Answer:

[tex]f=1.98\times 10^7\ Hz[/tex]

Explanation:

Given that,

The radius of circle, r = 0.53 m

The magnitude of the magnetic field, B = 1.3 T

We need to find the oscillator frequency. It is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

Put all the values,

[tex]f=\dfrac{1.6\times 10^{-19}\times 1.3}{2\pi \times 1.67\times 10^{-27}}\\\\f=1.98\times 10^7\ Hz[/tex]

So, the oscillator frequency is [tex]1.98\times 10^7\ Hz[/tex].

A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath. The temperature of the water bath is monitored, and it is determined from this data that 133.0 kJ of heat flows into the system during the reaction. The position of the piston is also monitored, and it is determined from the data that the piston does 241.0 kJ of work on the system during the reaction.

a. Does the temperature of the water bath go up or down?
b. Does the piston move in or out?
c. Does heat flow into or out of the gaseous mixture?
d. How much heat flows?

Answers

I feel like it would be B makes the most sense not sure tho

What does Boyle's Law state about the relationship between the pressure and volume of an ideal gas at constant temperature?
a) The product of pressure and volume increases as pressure decreases.
b) The sum of pressure and volume is constant.
c) The sum of product and volume decreases as volume increases.
d) The product of pressure and volume is constant.

Answers

Answer:

Option (d).

Explanation:

According to the Boyle's law, for a given mass of a gas, the pressure of the gas is inversely proportional to the volume of the gas keeping the temperature of the gas is constant.

So,

Let the pressure is P, volume is V and T is the absolute temperature of the gas.

Pressure proportional to the reciprocal of the volume.

[tex]P \alpha \frac{1}{V}\\\\P V = constant[/tex]

The correct option is (d).

Complete the following statement: The electromotive force is:______.a. the force that accelerates electrons through a wire when a battery is connected to it.b. the maximum potential difference between the terminals of a battery. c. the force that accelerates protons through a wire when a battery is connected to it.d. the maximum capacitance between the terminals of a battery.e. the potential difference between the terminals of a battery when the battery is not in use.

Answers

Answer:

The electromotive force is the maximum potential difference between the terminals of a battery.

The electromotive force is the maximum potential difference between the terminals of a battery.  The correct option is b.

What is electromotive force?

The electromotive force also called as EMF, is the force which causes current to flow from  the positive to negative terminal of the battery.

The electromotive force is the maximum potential difference between the terminals of a battery.

Thus, the correct option is b.

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No esporte coletivo, um dos principais fatores desenvolvidos é o desenvolvimento social. Qual desses não faz parte das virtudes ensinadas no esporte?

Companheirismo
Humildade
Ser justo (Fair Play)
Vencer independente do que precise ser feito

Answers

Answer:

fair palybtgshsisuehdh

Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off

Answers

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]

Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]

Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]

This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.

Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.

Required:
What is the magnitude of the dipole moment of the arrangement?

Answers

Answer:

The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ [tex]P=qd[/tex]

By putting the values, we get

⇒     [tex]=0.50\times 900[/tex]

⇒     [tex]=(0.50\times 10^{-9})\times 0.9[/tex]

⇒     [tex]=4.5\times 10^{-10} \ Cm[/tex]  

Answer:

The dipole moment is 4.5 x 10^-10 Cm.

Explanation:

Charge on each ball, q = 0.5 nC

Length, L = 900 mm = 0.9 m

The dipole moment is defined as the product of either charge and the distance between them.

It is a vector quantity and the direction is from negative charge to the positive charge.

The dipole moment is

[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g

Answers

Answer:

w = 1,066 rad / s

Explanation:

For this exercise we use Newton's second law

         F = m a

the centripetal acceleration is

         a = w² r

indicate that the force is the mass of the body times the acceleration

        F = m 0.58g = m 0.58 9.8

        F = 5.684 m

we substitute

       5.684 m = m w² r

       w = [tex]\sqrt{5.684/r}[/tex]

To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m

       w = [tex]\sqrt{ 5.684/5}[/tex]

       w = 1,066 rad / s

How does the theory of relativity explain the gravity exerted by massive objects?
A. More massive objects create stronger forces of gravity.
B. More massive objects create shallower curves of space-time.
C. More massive objects pull objects from farther away.
D. More massive objects create larger curves of space-time.

Answers

(D)

Explanation:

The more massive an object is, the greater is the curvature that they produce on the space-time around it.

The theory of relativity explain the gravity exerted by massive objects is

more massive objects create larger curves of space-time (option-d).

Do bigger objects exert more gravity?

The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.

Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.

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the speed of the pulse depends on what?​

Answers

Answer:

The pulse speed depends on the properties of the medium and not on the amplitude or pulse length of the pulse.

Explanation:

Hope this helps.

Shooting a rock with a slingshot converts
energy to__________,
energy____________,

Answers

Answer:

converts elastic energy to mechanical energy

Explanation:

Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude
(a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.

Answers

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 10.6 ft/s at point A and 15.6 ft/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.80 s to go from point B to point C. What is the cart's speed at point B

Answers

Answer:

a) [tex]a_{avg}=1.25ft/s^2[/tex]

b) [tex]v_b=13.35ft/s[/tex]

Explanation:

From the question we are told that:

Speed at point A [tex]v_A=10.6ft/s[/tex]

Speed at point C [tex]v_C=15.6ft/s[/tex]

Time from Point A to C [tex]T_{ac}=4.00s[/tex]

Time from Point B to C [tex]T_{bc}=1.80s[/tex]

Generally the equation for acceleration From A to B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{\triangle t}[/tex]

 [tex]a_{avg}=\frac{15.6-10.6}{4.0 }[/tex]

 [tex]a_{avg}=1.25ft/s^2[/tex]

Generally the equation for cart speed at B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{T_{bc}}[/tex]

 [tex]v_b=v_c-a_{avg}*T_{bc}[/tex]

 [tex]v_b=15.6ft/s-(1.25ft/s^2)(1.80)[/tex]

  [tex]v_b=13.35ft/s[/tex]

The working substance of a certain Carnot engine is 1.90 of an ideal
monatomic gas. During the isothermal expansion portion of this engine's
cycle, the volume of the gas doubles, while during the adiabatic expansion
the volume increases by a factor of 5.7. The work output of the engine is
930 in each cycle.
Compute the temperatures of the two reservoirs between which this engine
operates.

Answers

Answer:

Explanation:

The energy for an isothermal expansion can be computed as:

[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)

However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:

[tex]V_b = 2V_a[/tex]

Equation (1) can be written as:

[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]

Also, in a Carnot engine, the efficiency can be computed as:

[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]

[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]

In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]

relating the above two equations together, we have:

[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]

Making the work done (W) the subject:

[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]

From equation (1):

[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]

[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

If we consider the adiabatic expansion as well:

[tex]PV^y[/tex] = constant

i.e.

[tex]P_bV_b^y = P_cV_c^y[/tex]

From ideal gas PV = nRT

we can have:

[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]

[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]

From the question, let us recall  aw we are being informed that:

If the volumes changes by a factor = 5.7

Then, it implies that:

[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]

[tex]T_H = T_L (5.7)^{y-1}[/tex]

In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]

As such:

[tex]T_H = T_L (5.7)^{1.6-1}[/tex]

[tex]T_H = T_L (5.7)^{0.67}[/tex]

Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]

From in the question:

W = 930 J and the moles = 1.90

using 8.314 as constant

Then:

[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]

[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]

[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]

[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]

From [tex]T_H = T_L (5.7)^{0.67}[/tex]

[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]

[tex]\mathbf{T_H \simeq 125K}[/tex]

A uniformly charged thin rod of length L and positive charge Q lies along the x-axis with its left end at the origin as shown in Figure 1.

a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq

b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.

Answers

Answer:

b)

Explanation:

Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request

Answers

Answer:

Explanation:

Moment of inertia of solid sphere = 2/5 m R²

m is mass and R is radius of sphere.

Putting the values

Moment of inertia of solid sphere I₁

Moment of inertia of hollow  sphere I₂

Kinetic energy of solid sphere ( both linear and rotational )

= 1/2 ( m v² + I₁ ω²)                [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/5 m R² ω²)

= 1/2 ( m v² + 2/5 m v²)

=1/2 x 7 / 5 m v²

= 0.7 x 5 x 4² = 56 J .

This will be equal to work to be done to stop it.

Kinetic energy of hollow sphere ( both linear and rotational )

= 1/2 ( m v² + I₂ ω²)  [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/3 m R² ω²)

= 1/2 ( m v² + 2/3 m v²)

=1/2 x 5 / 3 m v²

= 0.833 x 5 x 4² = 66.64 J .

This will be equal to work to be done to stop it.

A scientist who studies fossils of ancient life forms .O ornithologist O Paleontologist O Ichthyologist O Marine Biologist .

Hurry !! First one to answer gets points !

Answers

Answer:

paleontologist

Explanation:

Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.

What is Fossil?

A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.

An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.

Therefore, the correct option is B.

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How are elastic and inelastic collisions different?


A: Elastic collisions occur when the colliding objects move separately after the collision; after inelastic collisions, the objects are connected and move together.

B: Elastic collisions occur when the objects are going the same direction when they collide; inelastic collisions occur when the objects are going in opposite directions when they collide.

C: Momentum is conserved in elastic collisions; momentum is not conserved in inelastic collisions.

D: Elastic collisions occur between objects of the same mass; inelastic collisions occur between different masses.

Answers

Answer:

a

Explanation:

Answer:

the answer is c

'

Explanation:

The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?

Answers

Answer:

a) the required angle in both radian and degree is  1.25 rad and 71.6°

b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Explanation:

Given the data in the question;

a)

we know that The expression for the angle subtended by an arc of circle at the center of the circle is,

θ = Length / radius

given that Length is 5000 miles and radius is 4000 miles

we substitute

θ = 5000 miles / 4000 miles

θ = 1.25 rad

Radian to Degree

θ = 1.25 rad × ( 180° / π rad )

θ =  71.6°

Therefore, required angle in both radian and degree is  1.25 rad and 71.6°

b)

The flight from Kampala to Singapore take 9 hours.

the plane's angular speed relative to the earth = ?

we know that, the relation between angular velocity and angular displacement is;

ω = θ / t

given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec

we substitute

ω = 1.25 rad / 32400 sec

ω = 3.86 × 10⁻⁵ rad/sec

Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic field for that wave at P

Answers

Answer:

[tex]6.63\times 10^8\ N/C[/tex]

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

[tex]\dfrac{E}{B}=c\\\\E=Bc[/tex]

Put all the values,

[tex]E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C[/tex]

So, the magnitude of the electric field is equal to [tex]6.63\times 10^8\ N/C[/tex].

Name the electrolyte in the chemical method of generating electricity​

Answers

Some of such commonly used compounds are Sodium Chloride, Nitric Acid, Sulphuric Acid, Sodium Acetate, Chloric acid, etc. The first battery was invented by Italian physicist Alessandro Volta in the year 1799 by generating continuous electric current using voltaic piles.

Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures

Answers

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off what

Answers

Answer:

Scattering is an interaction that can happen when a given particle or wave, like an electron, impacts a target or material. Then the electron changes it's original path and leaves some energy in the process. (This is a really simplified explanation of scattering, this is a really complex phenomenon, but let's not dive into that path)

Particularly, Davisson and Germer used a beam of electrons against a target of nickel, and these scattered electrons were detected by a detector. All of that in a vacuum chamber.

Then the correct answer is a nickel target.

"After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off a nickel target"

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces an amount of electric energy with the hot reservoir at 500 K during Day One and then produces the same amount of electric energy with the hot reservoir at 600 K during Day Two. The thermal pollution was:

Answers

Answer: hello your question lacks some vital information below is the complete question

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was

answer:

Total thermal pollution = 2.5 * 10^6 J

Explanation:

Low temperature reservoir = 300 K

hot reservoir temperature = 500 K

Electrical energy produced by plant ( W ) = 1 * 10^6 J

lets assume ; Q1 = energy absorbed , Q2 = energy emitted

W = Q1 - Q2  or  Q2 = Q1 - W  ( we will apply this as the formula for determining thermal pollution )

For day 1

T1 = 500k , T2 = 300k

applying Carnot engine formula

W / Q1 = 1 - T2/T1

∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J

thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J

for Day 2

T1 = 600k,  T2 = 300k

Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J

Thermal pollution; Q2 = Q1 - W  = 1 * 10^6 J

Therefore the Total thermal pollution =  1 * 10^6  + 1.5 * 10^6  = 2.5 * 10^6 J

A 300 kg block of dimensions 1.5 m × 1.0 m × 0.5 m lays on the table with its largest face.

Calculate:



Area of the largest face

Answers

Answer:

1.5

x 1.0

1.50

x 0.5

075.00

answer: 75.00m

Explanation:

I hope this help

The AM radio station WDRJ broadcasts news and sports at a frequency of 704 kHz (kilohertz). What is the wavelength of the radio waves this station broadcasts? _____ meters

Give your answer to the nearest hundredth of a meter (two places after the decimal). Just enter the number; do NOT use scientific notation.

Answers

Answer:

AM broadcasts occur on North American airwaves in the medium wave frequency range of 525 to 1705 kHz (known as the “standard broadcast band”). The band was expanded in the 1990s by adding nine channels from 1605 to 1705 kHz.

Part C – RC Circuits in AC Mode 1. Derive Equation 5-6 from Equation 5-5. 2. Using the τ’s you calculated and your measured resistance: a. Calculate the capacitances of the capacitors. b. Compare your calculated and measured values via percent error.

Answers

Answer: hi your question is incomplete attached below is the complete question

1) attached below

2) a)  31 Ω,   302.9 Ω

   b)  17.3 Ω ,  26.4 Ω

Explanation:

1) Deriving Eqn 5-6 from Eqn 5-5

attached below

2) using τ’s calculated and measured resistance

use given data

Tm1 = Rm1 * Cm1

       = 329.3 * 333 * 10^-9 = 109.65 μs

Tm2 = Rm2 * Cm2

       = 329.3 * 200 * 10^-9 = 658.6 μs

a) Capacitance of capacitors

For Cm1

   t₁₂ = 72 μs = R₁' Cm1 log²

 ∴ R₁' = ( 72 * 10^-6 ) / ( 333 * 10^-9 log² )

         = 31 Ω

For Cm2

 t₁₂ = 40 μs  , ∴  R₂' = 302.9 Ω

b) comparing calculated and measured values via percent error

errors

Rm1 - R₁' = 17.3 Ω

Rm2 - R₂' = 26.4 Ω

Other Questions
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