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
Explanation:
Here’s a revised version of the requirements;
Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.
A. Your weight including the saddle
C. Value of the rope's strength
B. Your weight
D. Value of the centripetal force
1 hour = 3,600 seconds
1 km = 1,000 meters
75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s
The mean speed during the deceleration is
(1/2)(20-5/6 + 0) = 10-5/12 m/s.
Traveling at this average speed for 21 seconds,
the bus covers
(10-5/12) × (21) = 218.75 meters.
C) B ⃗ = A ⃗ + C ⃗ Pay attention to the direction of the arrows (vectors).
The acceleration is calculated using the formula a= change in velocity/time. Here, it becomes A=10-0/2, simplifying to A=10/2, resulting in A=5 m/s².
