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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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An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai
rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion:        Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

                       Vf = Vi + a×t

                       Vf = 0 + 2.4×3.6

                       Vf = 2.4×3.6

                       Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0
3 years ago
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
3 years ago
In a pilot study to help train a group of data collectors who will evaluate childrens pain using a FACES rating scale, the resea
muminat

Answer:

Option c. Inter-rater Reliability

Explanation:

Here, the rating is done by a group of data collectors under training for evaluation children's pain on Faces scale which is a scale ranging from 0 to 10 with different expressions or faces with a happy face at 0 to a crying face at 10.

Also in Inter-rater Reliability, the relative consistency of a study or test is assessed and the extent to which different group members rated the same behavior, the consistency of which is evaluated.

Thus it can also be helpful in interviews, etc.

6 0
4 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
4 years ago
A 10 kg package is delivered to your house.
kolbaska11 [484]
In ur explanation make sure to use the terms
7 0
3 years ago
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