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VARVARA [1.3K]
3 years ago
9

A plane flies 1,480 miles in 4 hours, with the wind. On its return trip, it took 5 hours flying against the same wind to go 1,15

0 miles. How fast would the plane have flown without any wind? Type in your numerical answer only; do not type any words or letters with your answer.
Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:300 mi/hr

Explanation:

Given

Plane Flies 1480 miles in 4 hr with the wind

While returning it took 5 hr to cover 1150 miles against the wind

Let v and v_0 be the absolute speed of Plane and wind

Thus during with the flow of wind

time=\frac{Distance}{Speed}

\frac{1480}{v+v_0}=4

370=v+v_0   ------------1

During return trip wind speed oppose Plane thus

\frac{1150}{v-v_0}=5

230=v-v_0    ----------2

adding 1 & 2 we get

600=2v+v_0-v_0

v=300 mph

Thus Speed of plane without wind is 300 mi/hr

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A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
3 years ago
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Answer:

2N

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4-2=2

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3 years ago
The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will
inessss [21]

Answer:  weight on Jupiter = 869.75 N

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Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁  = 347,9 N

On the Jupiter,

G₂ = mg₂  

mass on the Earth = mass on the Jupiter !  

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

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3 years ago
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andriy [413]

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\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.

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<u>Given:</u>

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where,

\rm \Delta m,\ \Delta v are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.

The uncertainty in kinetic energy is given as:

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7 0
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