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
The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.
<span>(a) 0.0676 l
(b) 67.6 cc
We are given that 5.00 L of blood circulates through the heart every minute, with the heart beating 74.0 times per minute. Therefore, for each heartbeat, 5.00 L / 74.0 = 0.067567568 L of blood passes through. Rounding to three significant figures, the result is 0.0676 l. To convert liters to cubic centimeters, multiplying by 1000 leads to 67.6 cc of blood per heartbeat.</span>
