While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from a sodium street lamp, radio waves from an AM radio station, radio waves from an FM radio station, and microwaves from an antenna of a communications system. Rank these types of waves in terms of increasing photon energy, lowest first.

Answers

Answer 1
Answer:

Answer:

Explanation:

1. radio waves from am

2. radio waves from fm

3.yellow light from a sodium street lamp

4. microwaves from an antenna of a communications system.


Related Questions

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?
When the frequency of an electromagnetic wave increases, its energy A. increases. B. decreases. C. stays the same. D. It depends on the exact type of electromagnetic wave.
Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation.
determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.
If 49 jules of work is done when a 7 newton wagon of cheeseburgers is pulled, how far does it move?​

A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?

Answers

Answer: 1.64 *10^19 electrons

Explanation: In order to the explain this problem we have to consider the following:

The current= charge/time; so

as the electrons move in the tungsten wire we have:

0.526 C/s= N electrons per second* charge of electron=

N electrons/s= 0.526/1.6*10^-19= 3.28 *10^18 electrons/s

Then, during  5 seconds  will pass:

3.28 *10^18 electrons/s*5 5s= 1.64 *10^19 electrons

Answer:

1.64 x 10^19 electrons

Explanation:

The current is defined as I=ΔQ/Δt where ∆Q is the amount of charge flowing past a point in the filament. This charge is comprised of electrons that each carry charge of e = 1.602 × 10^-19 C. So ΔQ=Ne=IΔt and the number of electrons flowing through the filament in 5 s is N=IΔte=(0.526 A)(5 s)1.602×10^−19 C=1.64×10^19 electrons.

The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?​

Answers

Final answer:

The problem discusses the change in acceleration when a passenger is added to a car. It requires understanding of Newton's second law of motion, force equals mass times acceleration, and then recalculating the acceleration with the passenger added to the total mass.

Explanation:

This problem pertains to Newton's second law of motion, stating that the force applied on an object equals its mass times its acceleration (F = ma). Given that the initial acceleration of the Lamborghini Huracan with a driver is 0.80g or 0.80*9.80 m/s², we can calculate the force applied by the car. By multiplying the car's mass (1510 kg) with its acceleration, we will find the force.

Οnce we have the force, we can calculate the new acceleration if the 80 kg passenger rode along. Given that the force is constant, we determine the car's new acceleration by dividing this force with the new total mass (car mass + passenger's mass). So the question ultimately requires an application of the concepts of force, mass, and acceleration.

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Final answer:

The new acceleration of the Lamborghini Huracan with an added passenger can be calculated by finding the initial force using the car's mass and acceleration, and then using this force with the increased mass (original mass + passenger's mass) to find the new acceleration. The new acceleration will be less than the initial acceleration due to the increased mass.

Explanation:

To determine the new acceleration of the Lamborghini Huracan with an added passenger, we first calculate the initial force acting on the car. This can be done by using Newton's second law which states that Force = mass * acceleration. Initially, the acceleration is 0.80g (where g is acceleration due to gravity = 9.81 m/s²), and the mass is 1510 kg (including the driver). Therefore, the initial force = 1510 kg * 0.8 * 9.81 m/s².

However, when an 80-kg passenger rides along, the total mass becomes 1510 kg + 80 kg = 1590 kg. To find the new acceleration, we keep the force constant (as it is not affected by the introduction of the passenger) and rearrange the formula F = m*a as a = F/m. Use the increased mass to find the new acceleration. Please note that the new acceleration will be less than the initial acceleration due to increased mass.

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(25) A grinding machine is supported on an isolator that has two springs, each with stiffness of k and one viscous damper with damping constant of c=1.8 kNs/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. The floor oscillates with amplitude Y=3 mm and frequency of 18 Hz. Because of other design constraints, the stiffness of each spring must be greater than 3.25 MN/m. What is the minimum required stiffness of each of the two springs to limit the grinding machine’s steady-state amplitude of oscillation to at most 10 mm? Assume that the grinding machine and the wheel are a rigid body of weight 4200 N and can move in only the vertical direction (the springs deflect the same amount).

Answers

Answer:

k = 15.62 MN/m

Explanation:

Given:-

- The viscous damping constant, c = 1.8 KNs/m

- The floor oscillation magnitude, Yo = 3 mm

- The frequency of floor oscillation, f = 18 Hz.

- The combined weight of the grinding machine and the wheel, W = 4200 N

- Two springs of identical stiffness k are attached in parallel arrangement.

Constraints:-

- The stiffness k > 3.25 MN/m

- The grinding machine’s steady-state amplitude of oscillation to at most 10 mm. ( Xo ≤ 10 mm )

Find:-

What is the minimum required stiffness of each of the two springs as per the constraints given.

Solution:-

- The floor experiences some harmonic excitation due to the unbalanced engine running in the vicinity of the grinding wheel. The amplitude "Yo" and the frequency "f" of the floor excitation is given

- The floor is excited with a harmonic displacement of the form:

                         y ( t ) = Y_o*sin ( w*t )

Where,

           Yo : The amplitude of excitation = 3 mm

           w : The excited frequency = 2*π*f = 2*π*18 = 36π

- The harmonic excitation of the floor takes the form:

                       y ( t ) = 3*sin ( 36\pi *t )                          

- The equation of motion for the floor excitation of mass-spring-damper system is given as follows:

                      m*(d^2x)/(dt^2) + c*(dx)/(dt) + k_e_q*x = k_e_q*y(t) + c*(dy)/(dt)\n\n(m)/(k_e_q)*(d^2x)/(dt^2) + (c)/(k_e_q)*(dx)/(dt) + x = y(t) + (c)/(k_e_q)*(dy)/(dt)

Where,

     m: The combined mass of the rigid body ( wheel + grinding wheel body)        c : The viscous damping coefficient

     k_eq: The equivalent spring stiffness of the system ( parallel )

     x : The absolute motion of mass ( free vibration + excitation )

- We will use the following substitutions to determine the general form of the equation of motion:

                                   w_n = \sqrt{(k_e_q)/(m) } , \n\np = (c)/(2√(k_e_q*m) ) =  (1800)/(2√(k_e_q*428.135) ) =  (43.49628)/(√(k_e_q) )

Where,

               w_n: The natural frequency

               p = ζ = damping ratio = c / cc , damping constant/critical constant

                    

- The Equation of motion becomes:

                         (1)/(w^2_n)*(d^2x)/(dt^2) + (2*p)/(w_n)*(dx)/(dt) + x = y(t) + (2*p)/(w_n)*(dy)/(dt)

                     

- The steady solution of a damped mass-spring system is assumed to be take the form of harmonic excitation of floor i.e:

                         X_s_s = X_o*sin ( wt + \alpha  )       

Where,

              X_o : The amplitude of the steady-state vibration.

              α: The phase angle ( α )

- The steady state solution is independent from system's initial conditions and only depends on the system parameters and the base excitation conditions.

- The general amplitude ( X_o ) for a damped system is given by the relation:

                        X_o = Y_o*\sqrt{(1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2) }              

Where,

                r = Frequency ratio =  (w)/(w_n) =  \frac{36*\pi }{\sqrt{(k_e_q*g)/(W) } } = \frac{36*\pi }{\sqrt{(k_e_q)/(428.135) } } = (36*\pi*√(428.135)  )/(√(k_e_q) )

   

- We will use the one of the constraints given to limit the amplitude of steady state oscillation ( Xo ≤ 10 mm ):

- We will use the expression for steady state amplitude of oscillation ( Xo ) and determine a function of frequency ratio ( r ) and damping ratio ( ζ ):

              

                    ((X_o )/(Y_o))^2 \geq (1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2)\n\n((X_o )/(Y_o))^2 \geq (1+ ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135)  )/(√(k_e_q) ))^2)/(( 1 - ((36*\pi*√(428.135)  )/(√(k_e_q) ))^2)^2 + ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135)  )/(√(k_e_q) ))^2)\n\n

                    ((X_o )/(Y_o))^2 \geq ( 1 + (41442858448.85813)/(k_e_q^2 ))/([ 1 - ((5476277.91201  )/(k_e_q) )]^2 +  (41442858448.85813)/(k_e_q^2 )  )}\n\n((X_o )/(Y_o))^2 \geq ( (k_e_q^2 + 41442858448.85813)/(k^2_e_q ))/([ ((k_e_q - 5476277.91201)^2  )/(k_e_q^2) ] +  (41442858448.85813)/(k_e_q^2 )  )}\n

                   ((X_o )/(Y_o))^2 \geq ( k_e_q^2 + 41442858448.85813)/( (k_e_q - 5476277.91201)^2 +41442858448.85813 )}\n\n((10 )/(3))^2 \geq ( k_e_q^2 + 41442858448.85813)/( k^2_e_q  -10952555.82402*k_e_q +3.00311*10^1^3 )}\n\n\n10.11111*k^2_e_q  -121695064.71133*k_e_q +3.33637*10^1^4 \geq 0

- Solve the inequality (  quadratic ):

       k1_e_q \geq  7811740.790197058  (N)/(m)  \n\nk2_e_q \leq   4224034.972855095 (N)/(m)

- The equivalent stiffness of the system is due to the parallel arrangement of the identical springs:

                k_e_q = (k^2)/(2k) = (k)/(2)

- Therefore,

                  k1 \geq  7811740.790197058*2 = 15.62  (MN)/(m)  \n\nk2 \leq   4224034.972855095*2 = 8.448 (MN)/(m)  

- The minimum stiffness of spring is minimum of the two values:

                k = 15.62 MN/m

Madelin fires a bullet horizontally. The rifle is 1.4 meters above the ground. The bullet travels 168 meters horizont before it hits the ground. What speed did Madelin's bullet have when it exited the rifle?

Answers

The position vector of the bullet has components

x=v_0t

y=1.4\,\mathrm m-\frac g2t^2

The bullet hits the ground when y=0, which corresponds to time t:

1.4\,\mathrm m-\frac g2t^2=0\implies t=0.53\,\mathrm s

The bullet travels 168 m horizontally, which would require a muzzle velocity v_0 such that

168\,\mathrm m=v_0(0.53\,\mathrm s)

\implies v_0\approx320\,(\mathrm m)/(\mathrm s)

Final answer:

In the given physics problem, the bullet travels horizontally 168 meters before hitting the ground from a height of 1.4 meters. By calculating the time it takes for the bullet to fall to the ground due to gravity and then applying that time to the horizontal distance traveled, we find that the speed of the bullet when it exited the rifle was approximately 313.43 m/s.

Explanation:

The scenario defined is a classic Physics problem where an object is fired horizontally and falls to the ground due to gravity. We can calculate the horizontal speed of the bullet using the equations of motion associated with the vertical, free-fall motion of the bullet.

Gravity causes the bullet to fall to the ground. As we know that the height from the ground is 1.4 meters, we can calculate the time taken for the bullet to hit the ground using the equation: time = sqrt(2 * height / g), where g is the gravitational constant (approx. 9.8 m/s^2).

Substituting the given value, we get time = sqrt(2 * 1.4 / 9.8), which is around 0.536 seconds. The bullet travels 168 meters in this time horizontally, therefore its horizontal speed will be distance / time, which is 168 meters / 0.536 seconds = 313.43 m/s. So, Madelin's bullet had a speed of around 313.43 m/s when it exited the rifle.

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If the diameter of the black marble is 3.0cm, and bye using the formula for volume, what is a good approximation if it’s volume? Record to the ones place

Answers

Complete question is;

If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?

Answer:

14 cm³

Explanation:

We will assume that this black marble has the shape of a sphere from online sources.

Now, volume of a sphere is given by;

V = (4/3)πr³

We are given diameter = 3 cm

We know that radius = diameter/2

Thus; radius = 3/2 = 1.5 cm

So, volume = (4/3)π(1.5)³

Volume ≈ 14.14 cm³

A good approximation of its volume = 14 cm³

What are some of the benefits of learned optimism that have been found inresearch?
O
A. Fewer health problems
O
O
B. All of these
C. Making more money
O
D. A lower divorce rate

Answers

The benefits of learned optimism that have been found in research are Fewer health problems, Making more money, and a lower divorce rate. The correct option is B.

Learned optimismhas been associated with numerous benefits in research, including fewer health problems, making more money, and a lower divorce rate. Optimistic people tend to have better physical and mental health, which leads to fewer health problems. Additionally, optimistic people tend to be more successful in their careers and finances, which can lead to higher income and better financial stability. Finally, optimistic people tend to have better relationships, including lower divorce rates, as they are better able to handle conflicts and maintain positive attitudes toward their partners.

In summary, learned optimism has a range of benefitsfor individuals, including better physical and mental health, greater success in work and education, better relationships with others, and improved resilience. These benefits make learned optimism an important skill for individuals to develop in order to lead happier, healthier, and more successful lives.

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