So there are different ways this could be solved. I'll do try to explain it the way I was taught...
Simon is riding his bike at 12 km/hr relative to the sidewalk, away from where Keesha is.
Simon throws the ball at Keesha, at 5 km/hr.
Keesha sees the ball approaching her at (12-5) = 7 km/hr relative to the ground to her.
Therefore the answer is: 7 km/hr
<u>Answer:</u>
I think its
<em>10 kg</em>
plzz plzz plzz. mark as brainliest
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.
