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umka21 [38]
3 years ago
15

Relate temperature to the average kinetic energy in a material.

Physics
1 answer:
inn [45]3 years ago
5 0

<u>Answer:</u>

Temperature is always directly proportional to average kinetic energy of the material.

<u>Explanation:</u>

The average kinetic energy is given by the formula

\mathrm{K}=\frac{3}{2} \frac{R}{N A} T

Where K = average kinetic energy . R = gas constant , NA = avogadro's number, T = temperature

From the formula we can clearly see that K is directly proportional to T i.e kinetic energy increases with increase in temperature.

As a substance when it absorbs the heat , temperature increases which makes the faster movement of the particles thus increasing the kinetic energy.The above formula is applicable to gaseous molecules.

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Covalent bonds. Silicon, carbon, germanium, and a few other elements form covalently bonded solids. In these elements there are four electrons in the outer sp-shell, which is half filled. ... In the covalent bond an atom shares one valence (outer-shell) electron with each of its four nearest neighbour atoms.
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A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration​
san4es73 [151]

Answer:

0.4

Explanation:

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3 years ago
Les propriétés de l’air?
crimeas [40]

Answer:

Lorsque l'on détend l'air son volume augmente et sa pression diminue. L'air qui est un mélange de gaz est compressible et expansible. – Lorsque l'on comprime l'air, son volume diminue et sa pression augmente. – Lorsque l'on détend l'air, son volume augmente et sa pression diminue.

4 0
2 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
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Colt1911 [192]

Answer:

Photoelectric-type alarms aim a light source into a sensing chamber at an angle away from the sensor. Smoke enters the chamber, reflecting light onto the light sensor; triggering the alarm.

Explanation:

nfpa.org is the website with theanswer

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