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photoshop1234 [79]
3 years ago
9

He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a

ngle of θ = 50° from the vertical. He exerts a force perpendicular to the inclined wall. What is the magnitude of the force he exerts on the wall? Use g = 9.80 m/s^2

Physics
1 answer:
Aliun [14]3 years ago
4 0

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

\theta =50^{\circ}

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

F=mgcos\theta =76\times 9.8\times \sin 50

F=570.54 N

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Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

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Where:

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  • λ = de-Broglie wavelength
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<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

       \dfrac{m_\alpha}{m_p}=4

For the kinetic energies you find:

          \dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}

            \dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4

Thus:

           \dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}

          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

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