Answer:
The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.
Explanation:
Hi there!
The total momentum of the system is given by the sum of the momentum vectors of each cart. The momentum is calculated as follows:
p = m · v
Where:
p = momentum.
m = mass.
v = velocity.
Then, the momentum of the system will be the momentum of cart A plus the momentum of cart B (let´s consider the right as the positive direction):
mA · vA + mB · Vb
0.450 kg · 0.850 m/s + 0.300 kg · (- 1.12 m/s) = 0.047 kg · m/s
The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.
Answer:
The speed of the 1 kg red ball 8.04 m/s .
Explanation:
Given :
Separation between rods , d = 1.5 m .
Mass of the red ball is 1 kg .
Mass of the orange ball is 5.7 kg .
Angular velocity ,
.
Now , distance of center of mass from red ball is :

We know , speed is given by :

Hence , this is the required solution .
Explanation:
Could you also show the diagram
The sun will most likely illuminate Earth equally on both
poles during September and March. On these months, Earth will not be tilted
toward or away from the sun. This results to both poles receiving nearly the
same amount of light, which means that the length of days and nights are equal
all over Earth. This event is referred to as Equinoxes. During March, it is
called the Vernal Equinox, while during September, it is called the Autumnal
Equinox.
Answer:

Explanation:
<u>Center of Gravity
</u>
It refers to a point where all the forces of gravity of a body make a zero total torque. To find the solution we use the fact that the net force acting on the system boat-man is in every moment equal to zero. It's assured by the first Newton’s law, the center of gravity is at rest or in uniform motion in both moments. From an external viewer's point of view, the center of gravity remains unchanged. The formula to compute it is shown below

Originally, the man sits on the stern of the boat. His weight is applied at a distance xm=4.9 m from the pier (assumed as x=0). The boat is assumed to have a uniformly distributed mass applied at its center, i.e. at xb = 4.9 / 2 = 2.45 m. The center of gravity is located originally at


When the man walks to the prow, the boat moves x = 1.2 m from the pier, so its center is located at a distance

The man is located at

The center of gravity is computed now as


Both centers of gravity are equal, thus

Simplifying

Rearranging

Thus
