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2 years ago

The soccer field is _____.

Physics

2 answers:

Elan Coil [88]2 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]2 years ago

5 0

Modified for youth leagues

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All of the following are important reasons to take notes except.

jeka57 [31]

Its a waste of time, you have to not only write it down, but study it after too . other than that notes are great.

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3 years ago

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A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w

katovenus [111]

Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85 = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x 5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788 x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

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3 years ago

a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load distance=10 cm

We have to find the effort distance.

We know that

$load\times load\;distance=Effort\times effort\;distance$

Using the formula

$800\times 10=200\times effort\;distance$

Effort distance= $\frac{800\times 10}{200}$

Effort distance= $\frac{8000}{200}$

Effort distance=40 cm

Hence, the effort distance will be 40 cm.

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2 years ago

How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another

Harman [31]

Answer:

94,800 m

Explanation:

F = kq1 q2/r^2

1 = 9 x 10^9 x 1 / r^2

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1 year ago

Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x.

Contact [7]

The center of mass is given with this formula:
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$v=\frac{dv}{dt}$
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$\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\$
In our case it is:
$v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}$

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3 years ago