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3 years ago

The imaginary line from the tip of the football to the football to the sideline is called

Physics

1 answer:

pentagon [3]3 years ago

8 0

The imaginary line from the tip of the football to the football to the sideline is called Line of Scrimmage. I believe.

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Phương trình trạng thái tổng quát của khí lí tưởng diễn tả là

tatyana61 [14]

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pV/T=const goodluke

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3 years ago

Which feature or process is unique to nuclear power plants when compared to conventional coal-burning power plants?

Mazyrski [523]

Answer:

Control rods

Explanation:

Control rods is unique to nuclear power plants as they are mainly used in nuclear reactors to stabilize the rate of fission of plutonium or uranium. They are usually composed of chemical elements such as cadmium, silver, boron, and indium.

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3 years ago

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loris [4]

Answer:

I'm taking a wild guess at c

Explanation:

c. winter solstice

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3 years ago

What happens when an atom gives up loosely held valence electrons to another atom

vazorg [7]

Answer:

The number of valence electrons increases to 8 or, the atom gives up loosely held valence electrons.

Explanation:

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3 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago