He or she would enjoying doing a job as a caretaker
Hope this answer helps, cause Idk, I might be wrong, but I still, I used the correct formulas, so I might be correct
Let m₁ = 3.0 kg and v₁ = + 8 m/s (so right is positive), and m₂ = 1.0 kg and v₂ = 0. The total momentum of the two balls before and after collision is conserved, so
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' = + 5 m/s and v₂' are the velocities of the two balls after colliding, so
(3.0 kg) (8 m/s) = (3.0 kg) (5 m/s) + (1.0 kg) v₂'
Solve for v₂' :
24 kg•m/s = 15 kg•m/s + (1.0 kg) v₂'
(1.0 kg) v₂' = 9 kg•m/s
v₂' = (9 kg•m/s) / (1.0 kg)
v₂' = + 9 m/s
which is to say, the second ball is given a speed of 9 m/s to the right after colliding with the first ball.
Answer:
B. Gauss’s law
Explanation:
Gauss’s law is defined as the act of thinking something carefully, careful thought; meditation; deliberation. consideration acts like filter on the information a person receives, because it allows them to contemplate the implications of something before acting on it.
hope this helped!
Answer:
60 kg
80 kg
Explanation:
Work is equal to the change in energy.
W = ΔE = E − E₀
Let's start with block B. The work done by the tension force is equal to the change in energy. Initially, the block has potential energy. Finally, the block has kinetic energy.
W = ΔE
FΔy = ½ mv² − mgh
T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)
T (-2.0 m) = m (-2 m²/s²)
T = m (1 m/s²)
Now let's look at block A. The work done by tension and against friction is equal to the change in energy. Initially, the block has no energy. Finally, it has both kinetic and potential energy.
W = ΔE
Fd = ½ mv² + mgh − 0
(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)
(T − Nμ) (2.0 m) = 120 J
T − Nμ = 60 N
Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.
N = mg cos θ
N = (4.00 kg) (10 m/s²) (⅘)
N = 32 N
Substitute:
T − 32μ = 60 N
If μ = 0, then T = 60 N and m = 60 kg.
If μ = ⅝, then T = 80 N and m = 80 kg.
