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jeyben [28]
3 years ago
5

A certain carbon monoxide molecule consists of a carbon atom mc = 12 u and an oxygen atom mo = 17 u that are separated by a dist

ance of d = 128 pm, where "u" is an atomic unit of mass.
Part (a) write a symbolic equation for the location of the center of mass of the carbon monoxide molecule relative to the position of the oxygen atom. This expression should be in terms of the masses of the atoms and the distance between them. 50%

Part (b) Calculate the numeric value for the center of mass of carbon monoxide in units of pm. Grade Summary Deductions Potential 0% 100%
Physics
1 answer:
kvasek [131]3 years ago
7 0

Answer:

a)  x_{cm} = m₂/ (m₁ + m₂)   d , b)   x_{cm} = 52.97 pm

Explanation:

The expression for the center of mass is

                x_{cm} = 1 / M  ∑ x_{i} m_{i}

Where M is the total masses, mI and xi are the mass and position of each element of the system.

Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon

              x_{cm} = 1 / (m₁ + m₂)   (0+ m₂ x₂)

Let's reduce the magnitudes to the SI system

             m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg

             m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg

             d = 128 pm = 128 10⁻¹² m

The equation for the center of mass is

               x_{cm} = m₂/ (m₁ + m₂)   d

b) let's calculate the value

            x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷    128 10-12

            x_{cm} = 52.97 10⁻¹² m

            x_{cm} = 52.97 pm

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Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet
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Answer:

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Explanation:

We will apply the law of conservation of momentum here:

Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\

<u>v₂ = 5.7 m/s </u>

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A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie
jenyasd209 [6]

Answer:

0.5 A

Explanation:

N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm

The induced emf is given by

e = - N dФ/dt

Where, dФ/dt is the rate of change of magnetic flux.

Ф = B A

dФ/dt = A dB/dt

so,

e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V

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3 years ago
Particle-X has a speed of 0.720 c and a momentum of 4.350x1019 kgm/s. What is the mass of the particle? 2.0206 10-27 kg Hints: T
kirill115 [55]

Explanation:

Given that,

Speed of particle = 0.720 c

Momentum = 4.350\times10^{-19}\ kgm/s[/tex]

(I). We need to calculate the mass of the particle

Using formula of momentum

P=mv

m =\dfrac{P}{v}

m=\dfrac{4.350\times10^{-19}}{ 0.720\times3\times10^{8}}

m=2.013\times10^{-27}\ Kg

We need to calculate the rest mass of particle

Using formula of rest mass

m=\dfrac{m_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}

Where, m_{0} = rest mass

Put the value into the formula

m_{0}=2.013\times10^{-27}\times\sqrt{1-(\dfrac{0.720 c}{c})^2}

m_{0}=2.013\times10^{-27}\times\sqrt{1-(0.720)^2}

m_{0}=1.4\times10^{-27}\ kg

(b). We need to calculate the rest energy of the particle

Using formula of energy

E_{0}=m_{0}c^2

Put the value into the formula

E_{0}=1.4\times10^{-27}\times(3\times10^{8})^2

E_{0}=1.26\times10^{-10}\ J

(c).  We need to calculate the kinetic energy of the particle

Using formula of kinetic energy

K.E=mc^2-m_{0}c^2

K.E=(m-m_{0})\timesc^2

K.E=(2.013\times10^{-27}-1.4\times10^{-27})\times3\times10^{8}

K.E=1.84\times10^{-19}\ J

(d). We need to calculate the total energy of the particle

Using formula of energy

E=mc^2

Put the value into the formula

E=2.013\times10^{-27}\times(3\times10^{8})^2

E=1.812\times10^{-10}\ J

Hence, This is the required solution.

8 0
3 years ago
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