As we know that, molecular mass of ferric oxide, Fe2O3, is 159.69 grams.
Out of which, iron contributes 111.69 g (2 X 55.845 g) and oxygen contributes
48 g (3 X 16 g).
Each gram of iron (III) oxide contains 111.69/159.69 g of iron and 48/159.69
g of oxygen.
To produce 1000 g iron (III) oxide we need,
Iron = 111.69*1000/159.69 = 699.42 g
Oxygen = 48*1000/159.69 = 300.58 g
Answer:
v = 72.54 m/s
Explanation:
We have,
Length of a guitar string is 0.62 m
Frequency of a guitar string is 234 Hz
For guitar string,
The velocity of the wave in the string is given by :
So, the velocity of the waves in the string is 72.54 m/s.
Answer:
A) 80 N
Explanation:
The closer the particles get, the stronger the Coulomb force, which elongates choices C and D. The Coulomb force is inversely proportional to the distance squared. If the distance is cut in half, the force is multiplied by the reciprocal of (1/2)^2, which is 4. Multiplying it out, 20 times 4 is 80 N.
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
