Explanation:
Given that,
Initial speed of the bag, u = 7.3 m/s
Height above ground, s = 24 m
We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :
v = 22.88 m/s
So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.
The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down
Brown black fur and medium tail
By definition, the momentum is given by:
p = m * v
Where,
m = mass
v = speed.
On the other hand,
F = m * a
Where,
m = mass
a = acceleration:
For the boy we have:
p1 = m * v
p1 = (F / a) * v
p1 = ((710) / (9.81)) * (0.50)
p1 = 36.19 Kg * (m / s)
For the girl we have:
p2 = m * v
p2 = (F / a) * v
p2 = ((480) / (9.81)) * (v)
p2 = 48.93 * v Kg * (m / s)
Then, we have:
p1 + p2 = 0
36.19 + 48.93 * v = 0
Clearing v:
v = - (36.19) / (48.93)
v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
Answer:
the girl's velocity in m / s after they push off is -0.74 m / s
A. The molecules start packed together very tightly in a solid. Then when it turns to water, the molecules can move around each other freely, but still contained. When water turns to vapor, the molecules are going crazy moving around. They are not contained at all and bounce of of each other freely.
b. The temperature rises. (ice turns to water at 33 degrees and water turns to vapor at 212 degrees)
