Answer:
1.19cm^3 of glycerine
Explanation:
Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:
Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum
Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature
coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1
Change in temperature = 41-23 = 18oC
Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3
Answer: Yes Because it matches with the mass and the amount of force Hope this helps :>
Explanation:
When one body(sun) exerts a force on a second body(planet), the second body simultaneously exerts a force equal in magnitude and opposite in direction of the first body. Which makes the planet orbit in path C.
Hope this helps!!
Answer:
A & D
Explanation:
A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.
From the options, only options A & D fits this definition of single-displacement reactions.
For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.
For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.
The other options are not correct because they don't involve only and element and a compound on each side of the reaction.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
= K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4