<em>Introduction:</em>
The majority of high school students do not get enough sleep each night, and this trend has increased in recent years. A policy that is gaining traction is that of delaying school start time in order to allow students to sleep more hours each night.
<em>Body 1: </em>
Based on this assumption, the Seattle School District decided to delay school start time by nearly an hour during the 2016-2017 academic year. Instead of beginning the school day at 7:50, now students were able to begin the day at 8:45 am. This was done in order to counteract a trend that is common in industrialised societies. Adolescents operate best when they get nine hours of sleep. However, few students enjoy this.
<em>Body 2:</em>
According to the publication Science Advances, this change allowed students to enjoy 34 minutes of extra sleep on average each day. The total nightly sleep increased from six hours and 50 minutes to seven hours and 24 minutes. This meant that most students were sleeping longer than they were before. By sleeping longer hours, they were also more likely to be aware at school. It also reverses the trend on sleep deprivation in students.
<em>Body 3:</em>
I believe that this is a good change. The Seattle School District has proven that delaying school time does help students get more sleep. In turn, when students get closer to their recommended sleep time, they are more likely to perform better at school, feel less anxious, handle stress better and build positive relationships with people. I believe this is a change that should be implemented all over the country.
<em>Conclusion:</em>
In conclusion, evidence shows that delaying school start times can increase the time that students sleep, and in turn this can increase their overall health and well-being. This measure should be implemented all over the country.
Answer:
this question is incomplete
Answer:
Fluid mechanics has great application in important branches of science such as acoustics and aeroelasticity.
Explanation:
Acoustics studies the speed of transmission, propagation, storage and reproduction of sound. In industries, knowing how sound is transmitted and the speed in the different activities that are carried out, allow control programs and adjustments to be carried out to avoid damages to the employees.
Aeroelasticity comprises the study of three important forces; inertial forces, elastic forces, and aerodynamic forces.
This field of action of fluid mechanics is important because it allows space engineering to study or design new systems or aircraft that are stable in or outside the atmosphere.
Answer:
a. Heat Capacity = 1.756J/mol-K
b. Heat Capacity = 24.942J/mol-k
Explanation:
Given
Constant volume Cv = 0.81J/mol-k
T1 = 34K
Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K
First, The value of the temperature-independent constant.
Using Cv = AT³
Make A the subject of formula
A = Cv/T³
Substitute each values
A = 0.81/34³
A = 0.000020608589456543
A = 2.061 * 10^-5J/mol-k
The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature
Cv = AT³
So, The heat capacity when T = 44k is then calculated as
Cv = 2.061 * 10^-5 * 44³
Cv = 1.755522084266232
Cv = 1.756J/mol-K
(b) at 477 K.
Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:
Cv = 3R
Where R = universal gas constant
R = 8.314J/mol-k
Cv = 3 * 8.314
Cv = 24.942J/mol-k
Answer:
i) S–N plot is attached
ii) fatigue strength = 100 MPa
iii) fatigue life = 5.62 x 10^(5) cycles
Explanation:
i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)
ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.
So let's find the log of this and trace it on the graph attached.
Log(4 × 10^(6)) = 6.6
From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa
iii) The question says we should find the fatigue life for 120 MPa.
Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.
Thus,the fatigue life will be the inverse log of 5.75.
Thus, fatigue life = 10^(5.75)
Fatigue life = 5.62 x 10^(5)