Ba(NO₃)₂ + Na₂SO₄ → BaSO₄ + NaNO₃
By Trial and error.
Ba(NO₃)₂ + Na₂SO₄ → BaSO₄ + 2NaNO₃
Check:
LHS RHS
Ba 1 1
N 2 2
S 1 1
O 3*2 + 4 = 10 4 + 2*3 = 10
Using KE = 1/2mv^2
m = mass (80kg)
v = velocity (?)
KE = kinetic energy (2500J)
2500/((1/2) x (80)) = v^2
Square root the answer to get: 7.91 m/s
Answer:
Final velocity of the block = 2.40 m/s east.
Explanation:
Here momentum is conserved.
Initial momentum = Final momentum
Mass of bullet = 0.0140 kg
Consider east as positive.
Initial velocity of bullet = 205 m/s
Mass of Block = 1.8 kg
Initial velocity of block = 0 m/s
Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s
Final velocity of bullet = -103 m/s
We need to find final velocity of the block( u )
Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u
We have
2.87 = -1.442 + 1.8 u
u = 2.40 m/s
Final velocity of the block = 2.40 m/s east.