For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
Answer:
2406 miles
Explanation:
Let A be the starting position, B the junction position and C the final position after flying the 3.5 hrs. Also, let b be the distance from the starting point:

#Distance traveled in 1.5hrs is;

#Distance traveled in next two hrs:

#Now using the Cosine Rule:

Hence, the pilot is 2406 miles from her starting position.
Answer:
F - M a force exerted by scales on student
M a = M (9.8 + 4.9) m/s^2 upwards chosen as positive
a = 1.5 g net acceleration of student due to force of scales
W =M g weight of student (actual weight)
Wapp = M 1.5 * g apparent weight (on scales) of student
<span>Quarks are thought to be the basic component of protons and newtons.</span>
Answer:
14 m/s
Explanation:
u = 0, h = 10 m, g = 9.8 m/s^2
Use third equation of motion
v^2 = u^2 + 2 g h
Here, v be the velocity of ball as it just strikes with the ground
v^2 = 0 + 2 x 9.8 x 10
v^2 = 196
v = 14 m/s