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Kay [80]
3 years ago
6

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Oiled Steel fry

ing pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials (0.03), he quickly calculates the normal force. What is it (in N)?
Physics
1 answer:
Alinara [238K]3 years ago
5 0

Answer:

N  = 6.67 N

Explanation:

The frictional or frictional force is a force that arises from the contact of two bodies and opposes movement.

The friction is due to imperfections and roughness, mainly microscopic, that exist on the surfaces of the bodies. Upon contact, these roughnesses engage with each other making movement difficult. To minimize the effect of friction, either the surfaces are polished or lubricated, since the oil fills the imperfections, preventing them from snagging.

As the frictional force depends on the materials and the force exerted on one another, its magnitude is obtained by the following expression:

f = μ*N    Formula (1)

where:  

f is the friction force  (N)

μ is the coefficient of friction

N is the normal force (N)

Data

f = 0.2 N : frictional force between the steel spatula and the Oiled Steel frying pan

μ = 0.03 :coefficient of kinetic friction between the two materials

Calculating of normal force

We replace data in the formula (1)

f = μ*N  

0.2  = 0.03*N  

N  = 0.2 / 0.03

N  = 6.67 N

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The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
madam [21]

Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

  • E = hf  
  • E = h.v/λ
  • λ = h/(mv)
  • E = (1/2)mv²

Where:

  • E = kinetic energy of the particle
  • λ = de-Broglie wavelength
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  • v = speed of the particle
  • h = Planck constant

<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:

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            \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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During that period of time, the bird's displacement was 4 km east. So its velocity was (4km east)/(11hrs). That's 0.36 km/hour east. (rounded)
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