Answer:
The answer to the question is
3340800 m far
Explanation:
To solve the question, we note that acceleration = 29 m/s²
Time of acceleration = 8 minutes
Then if the shuttle starts from rest, we have
S = u·t+0.5·a·t² where u = 0 m/s = initial velocity
S = distance traveled, m
a = acceleration of the motion, m/s²
t = time of travel
S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far
<span>The amount of dissolved oxygen in water may decrease
because of the increase in organic matter in the water. <span>Aquatic organisms breathe and use oxygen. Large amounts of
oxygen are consumed by the decomposition of bacteria (when there are large
amounts of dead matter to decompose, there will be a significant number of
bacteria). Examples: dead organic matter (algae), wastewater, garden waste,
oils and fats, all this results in a decrease in dissolved oxygen in the water.</span></span>
Answer:
24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)
Explanation:
24) This problem can be solved by means of the following equation.

where:
DU = internal energy difference [J]
Q = Heat transfer [J]
W = work [J]
Since there are no temperature changes the internal energy change is equal to zero
DU = 0
therefore:

The work is equal to the heat transfered, W = 75 [J].
25) The heat transfer can be calculated by means of the following equation.
![Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]](https://tex.z-dn.net/?f=Q%20%3D%20m%2Ac_%7Bp%7D%2ADT%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%200.4%5Bkg%5D%5C%5Cc_%7Bp%7D%20%3D%20specific%20heat%20%3D%20897%5BJ%2Fkg%2AK%5D%5C%5CDT%3D%205%20%5BC%5D)
Q = 0.4*897*5 = 1794[J]
Work is equal to heat transfer, W = 1794[J]
26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

where:
m = mass = 0.5[kg]
g = gravity = 9.81[m/s^2]
h = 1.5 [m]
![E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D0.5%2A9.81%2A1.5%5C%5CE_%7Bp%7D%3D7.36%5BJ%5D)
The heat developed can be calculated by means of the following equation.
![Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]](https://tex.z-dn.net/?f=Q%3Dm%2Ac_%7Bp%7D%2ADT%5C%5CQ%3D0.5%2A130%2A1%5C%5CQ%3D65%5BJ%5D)
The number of times will be calculated as follows
n = 65/7.36
n = 8.8 (times) or 9 (times)
- m1=1500kg
- m_2=3000kg
- v_1=5m/s
- v_2=7m/s
Using law of conservation of momentum





4/3 m/s ( approximately 1.3333... m/s)