15 because i did the math, had this assignment, and also watch a video
Additionally, the coefficient of thermal expansion of the restorative material should be comparable to the coefficient of thermal expansion of the tooth structure, since a significant difference between the two could result in thermal-induced stress at the cavity wall and subsequent marginal failure.
This study's objective was to assess how thermal stress affected the marginal integrity of restorative materials with various adhesive and thermal characteristics. As an alternative to clinical trials, which are expensive and time-consuming, evaluation of restorative materials under laboratory simulations of clinical function is frequently carried out. Thermal cycling regimens, which are in vitro techniques that subject the restoration and the tooth to extremely high temperatures, are frequently used in laboratory simulations to replicate thermal stresses that naturally occur in vivo.
Learn more about Thermal Stress here-
brainly.com/question/20309377
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The sun is a yellow dwarf star, so the correct answer is yellow. Hope it helps! If you could vote me brainiest, that would be awesome!
Answer:
Explanation:
(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)
Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:
Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:
Ahora se reemplaza en la ecuación de período:
La aceleración experimentada por el planeta es:
Se reemplaza en la ecuación de período:
La distancia del planeta con respecto al sol es finalmente despejada:
![R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7BG%5Ccdot%20M_%7Bsun%7D%5Ccdot%20%5Cleft%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%20%5Cright%29%5E%7B2%7D%7D)
Finalmente, se sustituyen las variables y se determina la distancia:
![R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%5Cleft%286.674%5Ctimes%2010%5E%7B-11%7D%5C%2C%5Cfrac%7BN%5Ccdot%20m%5E%7B2%7D%7D%7Bkg%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%281.989%5Ctimes%2010%5E%7B30%7D%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cfrac%7B%2865%5C%2Ca%29%5Ccdot%20%5Cleft%28365%5C%2C%5Cfrac%7Bd%7D%7Ba%7D%20%5Cright%29%5Ccdot%20%5Cleft%2886400%5C%2C%5Cfrac%7Bs%7D%7Bd%7D%20%5Cright%29%7D%7B2%5Cpi%7D%20%5Cright%5D%5E%7B2%7D%7D)
Answer:
Explanation:
Electric field at the surface of the the lead 208 = KQ/ R²
where K = 8.99 × 10⁹ Nm² /C²
Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C
V of the lead = 208 v of a proton assuming they both are sphere
4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton
R³ = 208 r³
R = ∛( 208 r³) = 5.92r
replace r with 1.20 x 10-15 m
R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m
E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² = 0.233 × 10²² N/C = 2.33 × 10²¹ N/C
