Answer:
The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.
The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds
By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.
<h3>What are the limits?</h3>
First, we need to find the limits.
We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.
So if 58 in is the 100%, the 26% and 43% of that are:
- 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
- 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.
But we know that the CG is found to be 45.5% MAC, then it measures:
(45.5%/100%)*58in = 0.455*58in = 26.39 in
We need to compare it with the largest limit, so we get:
26.39 in - 24.94 in = 1.45 in
This means that the CG is 1.45 inches out of limits.
If you want to learn more about percentages, you can read:
brainly.com/question/14345924
Answer:
-0.045 N, they will attract each other
Explanation:
The strength of the electrostatic force exerted on a charge is given by
where
q is the magnitude of the charge
E is the electric field magnitude
In this problem,
(negative because inward)
So the strength of the electrostatic force is
Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.
Answer:
500 N
Explanation:
Natural bouyency will keep the diver in the same place, no matter what the mass of gfs. This means if the diver is displacing with a weight of 500 N, the upthrust will also be 500 N.
Answer:
x=4.06m
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)\\
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
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
