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koban [17]
3 years ago
5

a particle with charge Q is on the y axis a distance a from the origin and a particle with charge qi is on the x axis at a dista

nce d from the origin. The value of d for which the x component of the force on the second particle is the greatest is
Physics
1 answer:
vova2212 [387]3 years ago
3 0

Answer:

Explanation:

Force on Q due to q₁

= k Q.q₁ / r² , r is distance between two.

X component

= k Q.q₁ / r²  x cos θ ,  θ is angle r makes with x - axis.

= k Q.q₁ / r²  x cos θ

r = a / sinθ

X component

= k Q.q₁ sin²θ . cosθ/ a²  

= k Q.q₁ / 2 a² x sin2θ . sinθ

for maximum value of it ,

sin2θ = 1  , θ = 45°

a / d = tan45

a = d .

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Which factors affect heat transfer between a warm and a cool substance?
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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
kvasek [131]

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

7 0
3 years ago
Consider a piston filled with 3 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting
kap26 [50]

Answer: -5013.65\ J

Explanation:

Given

No of moles n=3

Temperature T=290\ K

Initial volume V_1=2\ m^3

Final volume V_2=1\ m^3

Work done in constant temperature process is

W=nRT\ln \left(\dfrac{V_2}{V_1}\right)

Insert the values

\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J

5 0
3 years ago
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