Explanation:
Calcium carbonate is a molecule that contains one atom of calcium, one atom of carbon, and three atoms of oxygen.
The answer is to this question is B
<h2>
Option 2 is the correct answer.</h2>
Explanation:
Elastic collision means kinetic energy and momentum are conserved.
Let the mass of object be m and M.
Initial velocity object 1 be u₁, object 2 be u₂
Final velocity object 1 be v₁, object 2 be v₂
Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s
Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M
Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J
Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M
We have
Initial momentum = Final momentum
24 = 3v₁ + 6M
v₁ + 2M = 8
v₁ = 8 - 2M
Initial kinetic energy = Final kinetic energy
96 = 1.5 v₁² + 18 M
v₁² + 12 M = 64
Substituting v₁ = 8 - 2M
(8 - 2M)² + 12 M = 64
64 - 32M + 4M² + 12 M = 64
4M² = 20 M
M = 5 kg
Option 2 is the correct answer.
Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
= 4 × 10^(-10)/2.5 × 10^2
= 1.6 × 10^(-12) volt
Answer:
64 J
Explanation:
The potential energy change of the spring ∆U = -W where W = work done by force, F.
Now W = ∫F.dx
So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)
∆U = - ∫-Fdx
= ∫F.dx
Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U = ∫₀²F.dx
= ∫₀²(40x - 6x²).dx
= ∫₀²(40xdx - 6x²dx)
= ∫₀²(40x²/2 - 6x³/3)
= ∫₀²(20x² - 2x³)
= [20x² - 2x³]₀²
= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)
= [(20(4) - 2(8)) - (0 - 0))
= [80 - 16 - 0]
= 64 J
