The three strata of study within the study of kinesiology are:
1. Core scientific domain: the core scientific domain deals with the sub disciplines that are found in kinesiology. Examples of these sub disciplines are biomechanics, sport medicine, exercise physiology, history of sport and dance, neurophysiology of performance, physiology of sport, sociology of sport, etc.
2. Sociocultural based form of movement: this aspect of kinesiology deals with the different types of games, sport and dance that are practised in different cultures and countries. The aspect studies expert skills that are involved in different sociocultural based movements. Examples of sociocultural based movements are aquatics, combative, individual sport, running, etc.
3. Methods in professional application: these deals with the professional training and careers that are available in kinesiology. There are generally five areas of professional training, they are: capacity building, management and administration, performance enhancement, instruction in physical education and research.
Answer:
<u>Acceleration is 2 m/s² </u>
<u>Distance</u><u> </u><u>is</u><u> </u><u>1</u><u>0</u><u>0</u><u> </u><u>m</u>
Explanation:
From definition of acceleration, Acceleration is the rate of change in velocity.
• Simplifying the definition, or modifying it;
• Let's formulate symbols:
- a is acceleration
- v is final velocity, v = 20 m/s
- u is initial velocity, u = 0 m/s [ at rest ]
- t is time, t = 10 seconds
Distance = ut + ½at²
Distance = (0 × 10) + (½ × 2 × 10²)
Distance = 0 + 10²
Distance = 100 meters
Answer:
1. Nitrogen
2. Oxygen
3. Carbon dioxide
4. Water vapor
5. Ozone
Explanation:
The atmosphere composes of 78% nitrogen which occupies the largest percentage followed by oxygen which takes up 21%, Argon takes up 1% then other components such as water vapor occupy between 0-7% and ozone takes 0.0-0.01. Moreover, 0.01-0.1 is occupied by carbon dioxide. Therefore, the answers for 1-5 are as follows.
1. Nitrogen
2. Oxygen
3. Carbon dioxide
4. Water vapor
5. Ozone
Answer:
Yes it is possible to increase the power with out changing the amount of work.
Explanation:
The power is defined by the amount of power divided by the time. This time is the one needed to do the work. We can understand this issue by analyzing an example with numeric values.
Work = 500 [J]
Time = 5 [s]
Power will be:
![Power=\frac{500}{5} \\Power=100 []watt]\\](https://tex.z-dn.net/?f=Power%3D%5Cfrac%7B500%7D%7B5%7D%20%5C%5CPower%3D100%20%5B%5Dwatt%5D%5C%5C)
Now if we change the time to 2 seconds:
![Power = 500 [J]/2[s]\\Power = 250 [watt]\\](https://tex.z-dn.net/?f=Power%20%3D%20500%20%5BJ%5D%2F2%5Bs%5D%5C%5CPower%20%3D%20250%20%5Bwatt%5D%5C%5C)
As we can see, the power was increased without the need to change the work.
Answer: d
Explanation:
mass of truck mt, kinetic energy of truck Ekt(Kt), Vt -speed of truck
mass of car ma, kinetic enregy of car Ekc(Kc), Vc-speed of car
mass of truck mt=2*mc
speed of truck Vt=2*Vc
Compare Kinetic energy of truck with kinetic energy of car using equation to calculatation ;
Kinetic energy of car:
Kc=(mc*Vc²)/2
Kinetic energy for truck:
Kt=(mt*Vt²)/2
Kt=((2*mc*(2*Vc)²)/2
Kt=(2*mc*4*Vc²)/2
Kt=8*mc*Vc²/2
Compare Kinteic energy truck with Kinetic energy of car:
Kt/Kc
Kt/Kc=((8*mc*Vc²)/2)/((mc*Vc²)/2)
Kt/Kc=8
