Answer:
the speed of the bullet before striking the block is 302.3 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 28.3 g = 0.0283 kg
mass of the wooden block, m₂ = 5004 g = 5.004 kg
initial velocity of the block, u₂ = 0
final velocity of the bullet-wood system, v = 1.7 m/s
let the initial velocity of the bullet before striking the block = u₁
Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.0283u₁ + 5.004 x 0 = 1.7(0.0283 + 5.004)
0.0283u₁ = 8.5549
u₁ = 8.5549 / 0.0283
u₁ = 302.3 m/s
Therefore, the speed of the bullet before striking the block is 302.3 m/s.
Answer:
v = √ 2e (V₂-V₁) / m
Explanation:
For this exercise we can use the conservation of the energy of the electron
At the highest point. Resting on the top plate
Em₀ = U = -e V₁
At the lowest point. Just before touching the bottom plate
Emf = K + U = ½ m v² - e V₂
Energy is conserved
Em₀ = Emf
-eV₁ = ½ m v² - e V₂
v = √ 2e (V₂-V₁) / m
Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron
D - tertiary consumer
This is because it is the farther up to food chain.
Answer:
1600 years half life
3200/1600 is 2 half lifes
first half life takes the sample to 1/2 it's original amount and the second takes it to 1/4 of it's original amount
so 1/4*7gm is 1.75gm
Explanation:
