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3 years ago

When a skier skis down a hill, the normal force exerted on the skier by the hill is

Physics

1 answer:

gayaneshka [121]3 years ago

5 0

Answer:

c. The normal force exerted on the skier by the hill is less than the weight of the skier.

Explanation:

Hi there!

Please see the attached figure for a better understanding of this explanation:

The vertical forces that act on the skier are the normal force (N) and the vertical component of the weight (wy) (see figure). Since the skier is not being accelerated in the vertical direction (perpendicular to the surface), the resulting net force acting in the vertical direction is zero:

∑Fy = 0

N - wy = 0

The vertical component of the weight is calculated using trigonometry of right triangles:

cos θ = adjacent / hypotenuse

cos θ = wy / W

W · cos θ = wy

Since:

cos θ < 1

Then:

W · cos θ < W

Then:

wy < W

and then:

The correct answer is "c": the normal force exerted on the skier by the hill is less than the weight of the skier.

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Find the de Broglie wavelength of the following. (a) a 4-MeV proton 14.3 Correct: Your answer is correct. fm (b) a 40-GeV electr

kherson [118]

a) de Broglie wavelength of a 4-MeV proton: 14.3 fm

b) de Broglie wavelength of a 40-GeV electron: 0.031 fm

Explanation:

The de Broglie wavelength of an object is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the particle

Here we want to find the de Broglie wavelength of a 4-MeV proton. The rest of mass of the proton in MeV is

$m_0 = 938 MeV$

And since $4MeV$ , this means that the proton is non-relativistic. So its kinetic energy is related to its momentum by

$E=\frac{p^2}{2m}$

which means

$p=\sqrt{2Em}$

where

$E=4 MeV \cdot 10^6 eV/MeV \cdot 1.6\cdot 10^{-19] J/eV=6.4\cdot 10^{-13} J$ is the kinetic energy

$m=1.67\cdot 10^{-27} kg$ is the proton mass

Substituting, we find

$\lambda=\frac{h}{\sqrt{2Em}}=\frac{6.63\cdot 10^{-34}}{\sqrt{2(6.4\cdot 10^{-13})(1.67\cdot 10^{-27})}}=14.3\cdot 10^{-15} m = 14.3 fm$

In this case, the electron has kinetic energy of 40 GeV, while the rest mass of an electron is

$m_0 = 0.511 MeV$

Since $40 GeV >> 0.511 MeV$ , the electron is ultra-relativistic: so we can rewrite its energy as

$E = pc$

The equation (1) can also be rewritten as

$\lambda = \frac{hc}{pc}$

where c is the speed of light. The quantity at the denominator is the energy, so

$\lambda=\frac{hc}{E}$

where:

$E=40 GeV = 40\cdot 10^9 eV \cdot (1.6\cdot 10^{-19})=6.4\cdot 10^{-9} J$ is the energy of the electron

And substituting, we find:

$\lambda=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{6.4\cdot 10^{-9}}=3.1 \cdot 10^{-17} m = 0.031 fm$

Learn more about de Broglie wavelength:

brainly.com/question/7047430

#LearnwithBrainly

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