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Viktor [21]
3 years ago
13

The soccer field is _____.

Physics
2 answers:
Elan Coil [88]3 years ago
8 0

Answer:

Modified for youth leagues

Explanation:

In 2015, USA developed a modified version of the known tackle football designed appropriately for youth leagues. This program is rooted in the idea that young players can take a gradual path to learning the sport: from flag football to modified tackle football to traditional 11-man tackle.

The modified youth league is known in America.

pychu [463]3 years ago
5 0
Modified for youth leagues

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6 0
3 years ago
Bob will not develop a good habit if he _____. works out for only one week develops a consistent schedule of exercise works out
Charra [1.4K]

Answer:

works out for only one week

Explanation:

In order to develop a good habit, there are certain things that you can do like:

-Develop a consistent schedule of exercise which allows you to work everyday over a period of time which helps you to create a routine that you would get used to it creating a habit.

-Work out with a friend which allows you to have someone to motivate you and that would go everyday with you which would help to be consistent.

-Use success strategies that are ways in which you can achieve your goal of creating a good habit to exercise.

According to this, the option that would not help Bob develop a good habit is if he works out for only one week as doing something everyday is what would help you to create the habit.

3 0
3 years ago
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
Usimov [2.4K]

Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

6 0
3 years ago
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