There are several many equations that are available to relate the distance,
speed, and time of a body moving vertically in gravity. Happily, the only one
I can always remember without looking it up happens to be the right one to
use for this question !
Distance = (1/2) x (gravity) x (time)²
3.8 m = (1/2) x (9.8 m/s²) x (time)²
Divide each side
by 4.9 m/s² :
(3.8 m) / (4.9 m/sec²) = (time)²
0.7755 sec² = time²
Square root
of each side:
0.88 second = time
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>
1. Pick a point on the top of the object and draw two incident rays traveling towards the mirror. Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the principal axis.
Answer:
0.5 m/s²
Explanation:
according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:
F = m×a
F for force
m for mass
a for acceleration
we use the given data and get:
20 = 40×a
we find a=20/40=0.5m/s²
