Physical activity can be a great way to connect with others and encounter a wide range of diversity in terms of age, gender, ethnicity, culture, and physical abilities.
What is encountered through physical activity?Physical activity is an excellent way to bring people from different backgrounds together, and it can be a great opportunity to celebrate diversity. I have heard stories of people who have participated in physical activities such as team sports, yoga classes, or running groups and have encountered a wide range of diversity in terms of age, gender, ethnicity, culture, and physical abilities.
For example, in team sports such as soccer or basketball, players may come from different cultural backgrounds and speak different languages. This diversity can create challenges in communication, but it can also provide an opportunity for players to learn about each other's cultures and develop a sense of community.
In yoga classes, participants may come from different age groups, genders, and physical abilities. The instructor may offer modifications to poses to accommodate participants with different needs, creating an inclusive and welcoming environment.
In running groups or races, participants may come from all walks of life and have different reasons for running. Some may be running for fitness, while others may be running to raise awareness for a cause or to honor a loved one. This diversity can create a supportive and motivating atmosphere, where runners can encourage each other and celebrate their achievements.
Therefore, physical activity can be a great way to connect with others, learn about different cultures and backgrounds, and celebrate diversity.
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Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 222 meters per second and the second car's velocity is 999 meters per second. At a certain instant, the first car is 888 meters from the intersection and the second car is 666 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)
Answer:
[tex]-7\ \text{m/s}[/tex]
Explanation:
x = Distance of 1st car = 8 m
y = Distance of second car = 6 m
[tex]\dfrac{dx}{dt}[/tex] = Velocity of 1st car = 2 m/s
[tex]\dfrac{dy}{dt}[/tex] = Velocity of 2nd car = 9 m/s
Distance between the cars at the given instant
[tex]r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{8^2+6^2}\\\Rightarrow r=10\ \text{m}[/tex]
From pythagoras theorem we have
[tex]r^2=x^2+y^2[/tex]
Differentiating with respect to time we get
[tex]2r\dfrac{dr}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}\\\Rightarrow \dfrac{dr}{dt}=\dfrac{x\dfrac{dx}{dt}+y\dfrac{dy}{dt}}{r}\\\Rightarrow \dfrac{dr}{dt}=\dfrac{-8\times 2-6\times 9}{10}\\\Rightarrow \dfrac{dr}{dt}=-7\ \text{m/s}[/tex]
The rate of change of the distance between the cars at the given instant is [tex]-7\ \text{m/s}[/tex].
Calculate the mass of an object given a net force of 5N right and an
acceleration of 5 m/s2 right.
Explanation:
F= MA
5N= M multiply by 5m/s 2
M=5/5
M = 1 kg