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3 years ago

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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.

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.

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If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net

nika2105 [10]

Answer:

2N

Explanation:

subtract rthe two forces to see which is greater

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

mass on Earth = mass on Jupiter = 35.5 Kg

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

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N = 869,75 N

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3 years ago

What inertia is present in streched rybber

MrRissso [65]

<h2>You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.</h2><h2>Hope it helps..</h2>

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Calculated the measurement uncertainty for Kinetic Energy when :mass = 1.3[kg] +/- 0.4[kg]velocity= 5.2 [m/s] +/- 0.2 [m/s]KE= 1

andriy [413]

Answer:

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

Explanation:

<u>Given:</u>

Mass, $\rm m\pm\Delta m = 1.3\pm 0.4\ kg.$

Velocity, $\rm v\pm \Delta v = 5.2\pm 0.2\ m/s.$

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:

$\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.$

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3 years ago