the higher concentration of molecules, the faster a reaction can occur
Answer: Chilling with my homies
Chicken wings, Chicken wings
Hotdog and baloney
Chicken and macaroni
Chillin' wit mah homiiieees
Chicken wing chicken wing
Hot dogs and balogna
Chicken and macaroni
Chilling with mah Chilling with mah
Chicken chicken wing chicken chicken wing
Chicken chicken chicken chicken chicken wing
Chicken chicken chicken chicken wing
Chicken wings, Chicken wings
Hotdog and baloney
Chicken and macaroni
Chillin' wit mah homiiieees
Chicken wings, Chicken wings
Hotdog and baloney
Chicken and macaroni
Chillin' wit mah homiiieees homiiieeees
Chicken wings, Chicken wings
Hotdog and baloney
Chicken and macaroni
Chillin' wit mah homiiieees
Chillin' wit mah
Explanation:
im sorry i had to do it now pls give me brainliest thank u and have a blessed day:)
Answer:
Change in electric potential energy ∆E = 365.72 kJ
Explanation:
Electric potential energy can be defined mathematically as:
E = kq1q2/r ....1
k = coulomb's constant = 9.0×10^9 N m^2/C^2
q1 = charge 1 = -2.1C
q2 = charge 2 = -5.0C
∆r = change in distance between the charges
r1 = 420km = 420000m
r2 = 160km = 160000m
From equation 1
∆E = kq1q2 (1/r2 -1/r1) ......2
Substituting the given values
∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)
∆E = 94.5 × 10^9 (3.87 × 10^-6) J
∆E = 365.72 × 10^3 J
∆E = 365.72 kJ
Answer: c
Explanation:
The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.
a. 2×2 atoms of Al ≠ 3×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!
b. 3×2 atoms of Al ≠ 3×1 atoms of Al
3×3 atoms of O ≠ 2×2 atoms of O
c. 2×2 atoms of Al = 4×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH ARE EQUAL, CORRECT ANSWER!!!
d. 2×2 atoms of Al ≠ 1×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
Answer:
Explanation:
one end of tank will be circular in shape . Area of circle A
= π r² , r is radius of the circle
= 3.14 x 3²
A = 28.26 ft³
To calculate force on the circular area , we first find pressure at the center of the circle which is at depth equal to r
pressure at the center = h d g ' here h = depth = r , d = density of milk
pressure = 3 x 64.6 x 32 poundal / ft²
= 6201.6 poundal / ft²
total force on circular face = pressure at the center x area of circle
= 6201.6 x 28.26
= 175257.21 poundal .
