Answer:
a) 
b) 
Explanation:
From the exercise we have <em><u>initial height and final X position</u></em>
a) From the concept of <u>free falling objects</u> we have that
Since the car runs off the edge of the table, that means the car is moving in x direction with 
and at the end of the motion 
<em><u>Solving for t </u></em>
t=± 0.6072 s
Since the time can not be negative, the answer is t=0.6072 s
b) To find the <u>horizontal velocity</u> of the car, we need to use the time that we just calculate
True lizards can stick to surfaces because their bulbous toes are covered in hundreds of tiny microscopic hairs called setae
Answer:
<em>J=600 kg m/s
</em>
Explanation:
<u>Impulse And Momentum
</u>
Suppose a particle is moving at a certain speed 
and changes it to 
. The impulse J is equivalent to the change of linear momentum. The momentum can be computed by
The initial and final momentums are given, respectively, by:
Thus, the change of momentum is
It's equal to the Impulse J
Our data is
Answer:
a)11.6m
b)45.55s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
a)
for this problem
Vo=0
Vf=319m/min=5.3m/s
a=1.2m/s^2
we can use the ecuation number 1 to calculate the time
t=(Vf-Vo)/a
t=(5.3-0)/1.2=4.4s
then we use the ecuation number 3 to calculate the distance
X=0.5at^2
X=0.5x1.2x4.4^2=11.6m
b)second part
We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)
we will have the distance traveled in with constant speed.
With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate
X=218-11.6x2=194.8m
X=VT
T=X/v
t=194.8/5.3=36.75s
Total time=36.75+2x4.4=45.55s
