Anatomy of a Wave worksheet can someone help me out with the answers????

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Answer 1

Answer: Part 1: here are the answers in order
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Please someone help, I’m very confused and it’s due soon, thanks

The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????

Answers

Answer:

The number of available energy states per unit volume is $4.01*10^(48)$

Explanation:

Given that,

Average energy $E=2*10^(-4)\ eV$

Photon = $4*10^(-5)\ eV$

We need to calculate the number of available energy states per unit volume

Using formula of energy

$g(\epsilon)d\epsilon=(8\pi E^2dE)/((hc)^3)$

Where, E = energy

h = Planck constant

c = speed of light

Put the value into the formula

$g(\epsilon)d\epsilon=(8*\pi*2*10^(-4)*4*10^(-5)*1.6*10^(-19))/((6.67*10^(-34)*3*10^(8))^3)$

$g(\epsilon)d\epsilon=4.01*10^(48)$

Hence, The number of available energy states per unit volume is $4.01*10^(48)$

Acetone, a component of some types of fingernail polish, has a boiling point of 56°C. What is its boiling point in units of kelvin? Report your answer to the correct number of significant figures.

Answers

Answer:

The boiling point of Acetone is 329K (in 3 significant figures)

Explanation:

Boiling point of Acetone = 56°C = 56 + 273K = 329K (in 3 significant figures)

Answer: using the formula 0°C + 273.15 = 273.15K the boiling point in units of kelvin to significant figures is 329.15k.

Explanation: The boiling point of a substance ( acetone) is the temperature at which the vapour pressure of the liquid substance equals the pressure surrounding it. The boiling point of acetone serves as it's characteristic physical properties. This is measured in degree Celsius (°C ) which can be converted to units of Fahrenheit or kelvin. To convert degree Celsius to kelvin this formula is used: 0°C + 273.15 = 273.15K . Given that acetone has boiling point of 56°C,from the formula 0°C is substituted for 56°C. This gives us:

56°C + 273.15= 319.15k.

Also,measurements given in Kelvin will always be larger numbers than in Celsius and the Kelvin temperature scale does not use the degree (°) symbol because Kelvin is an absolute scale, based on absolute zero, while the zero on the Celsius scale is based on the properties of water. I hope this helps. Thanks

Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , measured relative to the merry-go-round. neglect friction and the size of each child.

Answers

by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

$m*2*R + Iw = 0$

$w = -(2mR)/(I)$

so merry go round will turn in opposite direction with above speed

A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power

Answers

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.

Answers

Answer:

av=0.333m/s, U=3.3466J

$v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_(B2)-v_(A2)$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

$2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s$

John, who has a mass of 65kg stands at rest on the ice. He catches a 10kg ball that is thrown to him at 5m/s.

Answers

The momentum of John after catching the ball is 50 kg.m/s.

"Your question is not complete, it seems to be missing the following information";

find John's momentum

The given parameters;

mass of John, m = 65 kg
mass of the ball caught by John, m' = 10 kg
initial velocity of John, u = 0
initial velocity of the ball, v = 5 m/s

Apply the principles of conservation of linear momentum to determine the momentum of John.

The momentum of John is calculated as follows;

P = mu + mv

P = (65 x 0) + (10 x 5)

P = 0 + 50

P = 50 kg.m/s

Thus, the momentum of John after catching the ball is 50 kg.m/s.

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A local meteorologist reports the day’s weather. "Currently sunny outside, 34°F. Skies will become overcast later this afternoon, as temperatures drop to 25°F, with windy conditions out of the north at 10–15 miles per hour. Radar indicates 2–3 inches of snow expected to fall later tonight.” Which information is qualitative? These are non-numerical, descriptive data. These are numerical data that have been measured. “sunny” “25°F” “2–3 inches of snow” “10–15 miles per hour”

What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)

The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference. b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference. c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference. d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.