(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
<h3 /><h3>The given parameters:</h3>
- Mass of block 1 = m1
- Mass of block 2, = m2
- Height of block 1 above the ground, = h1
- Height of block 2 above the ground = h2
The total initial mechanical energy of the two block system is calculated as follows;

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

Thus, we can conclude the following before the block m2 reaches the ground;
- For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
- For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
Learn more about conservation of mechanical energy here: brainly.com/question/332163
Answer:
The correct answer is b. Differences in Path Lenght
Explanation:
In this area of knowledge, there two main ways of referring to Path Lenght:
1. The total distance an object travels from a starting point, without considering trajectory.
2. When talking about wavelength, because this relates to light and how it travels throughout different mediums. This concept is used in calculating constructive and destructive interference of waves, and is responsible for seeing different shades of tea color.
Hope this helps!
Given that in a parallel circuit:
R1 = 12 ohms
R2= 15 ohms
I = 12 A
I2 = 4 A
V=?
R=?
R3 =?
P=?
Since,
V= IR
or,
V2 = I2 * R2
V2= 4* 15
V2 = 60V
Since in a parallel circuit voltage remain same in all component of the circuit and is equal to the source voltage.
Therefore,
V= V1 = V2 = V3 = 60V
Since,
V= IR
R= V/I
R= 60/12
R= 5 ohm
That is total resistance is equal to 5 ohms.
Since for parallel circuit,
1/R= 1/R1 + 1/R2 + 1/R3
1/5= 1/12+ 1/15 + 1/R3
or
1/R3= 1/5- 1/12- 1/15
1/R3= 1/20
or
R3= 20 ohms
Since,
V=IR
I= V/R
I1= V1/ R1
I1= 60/12
I1= 5 A
I3= V3/R3
I3= 60/20
I3= 3A
Since,
P=VI
P= 60*12
P= 720 watt
P1= V1* I1
P1= 60* 5
P1= 300 watt
P2= V2* I2
P2= 60* 4
P2= 240watt
P3= V3*I3
P3= 60*3
P3= 180 watt
Hence we have,
R1= 12 ohms , R2= 15 ohms, R3= 20 ohms, R= 5 ohms
I1= 5A, I2= 4A, I3= 3A, I= 12 A
V1= V2= V3= V= 60V
P1= 300 watt, P2= 240 watt, P3 = 180 watt, P= 720 watt
Answer: Trotting uses only 75 percent of the energy as galloping
Explanation: Trotting is only 300 J/m, whereas galloping is roughly 400 J/m