<span>A similar thing occurred with the circles as did with the electroscope. When we initially brought the charged pole close to, the bar pulled in the circle since it was of polarization. At that point, once the bar touched the circle, the pole repulsed the circle. This is on the grounds that once the pole and circle touched, the electrons exchanged thus did the protons, consequently leaving the circle with a positive net charge. The nearer the bar is to the circle the more it repulsed, however, it didn't influence the charge of the circle once the circle was touched by the pole.</span>
By Gay Lussacs law you can find the pressure. First both temperatures of Celsius must change to Kelvin by adding 273. Temperature one will be 308K and temperature 2 will be 258K
With this info, you can now find the pressure with Lussacs law
P1 = P2
— —
T1 T2
Pressure 1 is given which is 32 psi so just plug it all in and find P2
32 = x
—— ——
308 258
308x = 8256 (Cross multiply)
X = 26.8 (divide both sides by 308)
Answer is 26.8 PSI
This makes sense because as temperature increases pressure increases, as well as when temperature decreases, pressure decreases. Since it’s a colder day the pressure will be lower.
Container which is heated
Answer:
yes
Explanation:
oak trees don't have vertebrates
Answer and Explanation:
a) The direction is shown in the cube diagram attached to this solution.
b) the angle between two planes (h₁, k₁, l₁) and (h₂, k₂, l₂) is given by the formula,
Cos Φ = (h₁h₂ + k₁k₂ + l₁)/√((h₁² + k₁² + l₁²)(h₂² + k₂² + l₂²))
For (111) and (112)
Cos Φ = (1.1 + 1.1 + 1.2)/√((1² + 1² + 1²)(1² + 1² + 2²))
Cos Φ = (1 + 1 + 2)/√((1+1+1)(1+1+4))
Cos Φ = 4/√(3×6)
Cos Φ = 4/√18
Φ = cos⁻¹ (4/√18) = 19.56°
c) equation 3.3 is missing from the question, I would be back to provide the answers to that as soon as the equation is provided!
Hope this Helps!!
