Answer:
Explanation:
According to the parameters provided,
mass of the clay lump, m₁ = 0.05 kg
initial velocity of the lump, u₁ = 12 m/s
mass of the cart, m₂ = 0.15 kg
initial speed of the cart, u₂ = 0
As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:
The resultant speed is v = 3 m/s.
Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.
This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
The final mass will be slightly lower due to evaporation. I learned this back in third grade, so it's surprising you're in high school and don't know this.
A 40 kg child throws a 0.5 kg stone at a velocity of 5 m/s. To find the recoil, we apply the conservation of momentum formula: m1•v1 + m2•v2 = 0, where m1 is the mass of the child, and v1 is the child's recoil velocity. Applying the known values results in 40•v1 = -0.5 × 5, leading to v1 = -2.5 / 40, which simplifies to v1 = -0.0625 m/s. Thus, the child's recoil speed is 0.0625 m/s.
