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3 years ago

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You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

hat have lost their brakes on mountain grades. Typically, such a lane is horizontal (if possible) and about 36.0 m36.0 m long. Think of the ground as exerting a frictional drag force on the truck. A truck enters a typical runaway lane with a speed of 50.5 mph50.5 mph ( 22.6 m/s22.6 m/s ). Use the work-energy theorem to find the minimum coefficient of kinetic friction between the truck and the lane to be able to stop the truck.

1 answer:

kati45 [8]3 years ago

6 0

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

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1 year ago

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2 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

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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, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

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From the conservation of momentum,

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

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

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$P=h/\lambda,$

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

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$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

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3 years ago

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Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

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Generally converting $T_3$ to Fahrenheit

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=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

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