PHYSICS COLLEGE

A businessperson took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours, and each way the trip was 200 miles. What was the speed of the wind that affected the plane, which was flying at a speed of 120mph? Round your answer to the nearest whole number.

Answers

Answer 1

Answer:

Answer:

Speed of the wind is 48.989 mph

Explanation:

We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind = $(200)/(120+x)$

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind $=(200)/(120-x)$

Total time is given as t= 4 hour

So $(200)/(120+x)+(200)/(120-x)=4$

$(200(120-x)+200(120+x))/((120+x)(120-x))=4$

$48000=57600-4x^2$

$4x^2=9600$

x = 48.989 mph

Answer 2

Answer:

Answer:

Explanation:

Type Distance Rate Time

Headwind 200 120-r 200/120-r

Tailwind 200 120 - r 200/120 - r

We know the times add to 4, so we write the equation:

200/120−r + 200/120 + r = 4

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)

200(120−r)+200(120+r)=4(120−r)(120+r)

Factor the 200 and simplify inside the parentheses to find:

200(120−r+120+r)=4(1202−r2)

200(240)=4(1202−r2)

200(60)=120^2−r^2

12,000=14,400−r^2

−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.

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The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.

Answers

Answer:

6052114.67492 m

$12.742* 10^(6)\ m$

Explanation:

v = Velocity of cosmic ray = 0.88c

c = Speed of light = $3* 10^8\ m/s$

d = Width of Earth = Diameter of Earth = $12.742* 10^(6)\ m$

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

$d_e=d\sqrt{1-(v^2)/(c^2)}\n\Rightarrow d_e=12.742* 10^(6)\sqrt{1-(0.88^2c^2)/(c^2)}\n\Rightarrow d_e=6052114.67492\ m$

The Earth's width is 6052114.67492 m

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

Hence, the width is $12.742* 10^(6)\ m$

Two loudspeakers, 4.0 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m.a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Answers

Answer:

a) 343.0 Hz b) 686.0 Hz

Explanation:

a) First, we need to know the distance to both speakers.

If the person is at halfway between the two speakers, and they are 4.0 m apart, this means that he is at 2.0 m from each speaker.

So, if he moves 0.25 m towards one of them, the distance from any speaker will be as follows:

d₁ = 2.0 m-0.25 m= 1.75 m

d₂ = 2.0 m + 0.25 m = 2.25 m

The difference between these distances is the path difference between the sound from both speakers:

d = d₂ - d₁ = 2.25 m - 1.75 m = 0.5 m

If the person encounters at this path difference a minimum of sound intensity, this means that this distance must be an odd multiple of the semi-wavelength:

d = (2*n-1)*(λ/2) = 0.5 m

The minimum distance is for n=1:

⇒ λ = 2* 0.5 m = 1 m

In any wave, there exists a fixed relationship between the speed (in this case the speed of sound), the wavelength and the frequency, as follows:

v = λ*f, where v= 343 m/s and λ=1 m.

Solving for f, we have:

$f =(343.0 m/s)/(1.0 m) = 343 Hz$

b) If the person remains at the same point, for this point be a maximum of sound intensity, now the path difference (that it has not changed) must be equal to an even multiple of the semi-wavelength, which means that it must be met the following condition:

d = 0.5 m = 2n*(λ/2) = λ (for n=1)

if the speed remains the same (343 m/s) we can find the new frequency as follows:

$f =(v)/(d) =(343 m/s)/(0.5m) =686.0 Hz$

⇒ f = 686.0 Hz

Final answer:

Two speakers create peaks and troughs of sound intensity due to constructive and destructive interference of waves. Using wave properties, the frequency of the sound when a minimum intensity is experienced 0.25m from the center is 680Hz. Increasing the frequency, the first to produce maximum intensity at the same position is about 2720Hz.

Explanation:

The behavior of sound intensity in this question is due to wave interference, specifically, constructive and destructive interference of sound waves. When you stand halfway between the speakers, the sound waves from each speaker are in phase, which means the pressure variations combine to create an intensified sound, known as constructive interference.

When you move towards one of the speakers and detect a minimum in sound intensity, this is due to destructive interference, which occurs when the crest of one wave overlaps with the trough of another, canceling each other and producing a minimum sound level.

a. The frequency of the sound can be calculated using the formula for wave speed, v = f.lambda, where v is the speed of sound (340 m/s under normal conditions), f is the frequency, and lambda is the wavelength. In this case, a minimum sound intensity indicates one-half wavelength. So, lambda = 0.5 m. Thus, frequency, f = v/lambda = 340/0.5 ~ 680 Hz.

b. When you increase the frequency while remaining 0.25m from the center, the first frequency for which the location will be a maximum of sound intensity will be when you are an integral multiple of the wavelength away from the source. Thus if we let this be 2λ, we can calculate the frequency as f = v / λ = v / (0.25m / 2) = 340 / 0.125 ~ 2720 Hz.

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Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. express your answer numerically in seconds. neglect air resistance.

Answers

It will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

We can assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. We also assume that air resistance can be neglected.

Assuming that the package was dropped from rest at a height of h, the time it takes for the package to reach sea level can be calculated using the equation:

h = (1/2) * g * t²

where g is the acceleration due to gravity (9.8 m/s²) and t is the time it takes for the package to reach sea level.

Solving for t, we get:

t = sqrt(2h/g)

To convert the initial velocity of the package from km/hour to m/s, we can use the conversion factor:

1 km/hour = 0.2778 m/s

Therefore, the initial velocity of the package is:

v0 = 342 km/hour * 0.2778 m/s/km/hour = 95.0 m/s

if the package was dropped from a height of 5000 meters, the time it takes for the package to reach sea level is:

h = 5000 m

t = sqrt(2h/g) = sqrt(2*5000/9.8) = 32.0 seconds

Therefore, it will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

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Final answer:

The time a dropped package takes to reach sea level from a plane is determined by its vertical motion. If the package retains only horizontal velocity when released, the time taken would be calculated using the height from which the object is dropped. However, to give a numerical value of time, we need to know the exact height.

Explanation:

The time it takes for the package dropped from the plane to reach sea level is determined exclusively by the package's vertical motion, assuming the package does not face air resistance. Specifically, the time of flight for a projectile launched and landing at the same elevation is governed by the equation: t = 2*v/g, where v represents the initial vertical velocity and g is the acceleration due to gravity. From the scenario, it seems the package retains only horizontal velocity when released since it's dropped down directly rather than being thrown downward, hence rendering initial vertical velocity as zero. Simply put, the package only begins to accelerate in the vertical direction once it's dropped, meaning the time taken would be calculated using the equation: t = √(2h/g), h being the height from which the object is dropped.

In the provided context, unfortunately, we need the height from which the package is dropped to give a specific numerical value of the time in seconds. If we knew the height of the plane at the time the package was dropped, we'd recalculate the time in seconds more precisely.

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What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answers

Answer:

$\lambda=1282nm$

Explanation:

The wavelength of the photons emitted due to an atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

$(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2)})$

Here $R_H$ is the Rydberg constant for hydrogen and $n_1,n_2$ are the lower and higher quantum number for the energy levels of the atomic electron transition, respectively. Replacing the given values and solving for $\lambda$

$(1)/(\lambda)=1.097*10^7m^(-1)((1)/(3^2)-(1)/(5^2)})\n(1)/(\lambda)=7.81*10^5m^(-1)\n\lambda=(1)/(7.81*10^5m^(-1))\n\lambda=1.282*10^(-6)m\n\lambda=1.282*10^(-6)m*(1nm)/(10^(-9)m)\n\lambda=1282nm$

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answers

To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$