Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law;

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

Answer: damping coefficient = 1.5×10^5Ns/m

Explanation:

Refer to the attached file for the solution

Answer: 592.37m

Explanation:

Person D is represented by the blue line.

The total displacement is calculated by subtracting the initial position from the final position. Starting at (0,0), the path consists of moving down two blocks, then right six blocks, followed by moving up four blocks, and finally left one block.

Here, I consider the positive direction of the x-axis to the right and the positive direction of the y-axis as upward.

Thus, the new coordinates will be, with B representing a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

Given that B = 110m

P = (550m, 220m)

The displacement corresponds to the length of the vector, since the change from the initial position (0,0) to P is simply P:

P = √(550^2 + 220^2) = 592.37m