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2 years ago

Which levels of government establish and implement educational requirements for minors

Physics

1 answer:

romanna [79]2 years ago

5 0

Answer:

Federal Role in Education is primarily a State and local responsibility in the United States. It is States and communities, as well as public and private organizations of all kinds, that establish schools and colleges, develop curricula, and determine requirements for enrollment and graduation.

Explanation:

The Role of Public Education in Supporting American Democracy. Since the founding of public education in the United States, public schools have been charged not only with giving future workers skills for the private marketplace, but also with preparing students to be citizens in a democracy.

You might be interested in

_____ replacement involves one element replacing another element in a compound.

Scorpion4ik [409]

Answer:

Single replacement

Explanation:

A single replacement reaction is a reaction in which the less reactive elements are replaced by the more reactive element in a compound. This is so because the more reactive element are higher in the electrochemical series element than the less reactive elements

3 0

2 years ago

Ellus

Lesechka [4]

Answer:

W'=125.44 N

Explanation:

The weight of a person on the surface of Earth is 784 N

Weight is given by :

W = mg

m is mass of the person and g is acceleration due to gravity on the surface of Earth (10 m/s²)

$m=\dfrac{W}{g}\\\\m=\dfrac{784}{10}\\\\m=78.4\ kg$

The acceleration due to gravity on the surface of Moon, g' = 1.6 m/s²

Weight of the person on the moon is :

W'=mg'

$W'=78.4\ kg\times 1.6\ m/s^2\\\\W'=125.44\ N$

Hence, the person would weigh 125.44 N on the Moon.

6 0

2 years ago

Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .