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Sergio039 [100]
2 years ago
8

Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is direct

ly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied
Physics
1 answer:
svlad2 [7]2 years ago
6 0

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

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Answer:

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Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

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\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}} (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

T_{3}=0.8T_{2} (4)

  • For Isochoric process:

\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}} (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

P_{4}=1.15P_{3} (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

P_{4}=1.15*1.1P_{1}=1.265P1

This is, the new pressure of the gas increases by 26.5%  with respect to initial pressure.

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V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1

so the final volume decreases 27% with respect to initial volume.

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T_{4}=1.15*0.8T_{1}=0.92T1

The new temperature decreases 8% with respect to initial volume.

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Why are images reflected from a rough surface not as clear as those reflected from a smooth
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Just as images are reflected from the surface of a mirror, light reflected from a smooth water surface also produced a clear image. ... Consequently, the outgoing rays are reflected at many different angles and the image is disrupted. Reflection from such a rough surface is called diffuse reflection and appears matte.

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An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t
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Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

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p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f)  in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

6 0
3 years ago
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