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3 years ago

12. What does positive and negative acceleration indicate?

Physics

1 answer:

Vesnalui [34]3 years ago

4 0

Answer:

Correct answer: positive acceleration indicates accelerated movement (motion) and negative acceleration indicates slow movement (motion)

Explanation:

v = v₀ + a t accelerated movement

v = v₀ - a t slow movement

God is with you!!!

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Water flowing from a waterfall before it hits the pond below?

anastassius [24]

Answer:

Kinetic Energy

Explanation:

Ang prinsipyo ay nagsasaad na ang enerhiya ay hindi maaaring malikha o masira, ngunit maaari lamang ma-convert mula sa isang anyo patungo sa isa pa. Ang tubig sa tuktok ng napakataas na talon ay nagtataglay ng gravitational potential energy. Habang bumabagsak ang tubig, ang enerhiya na ito ay na-convert sa kinetic energy, na nagreresulta sa isang daloy sa isang mataas na bilis.

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2 years ago

Can u find the . in this?

nikklg [1K]

Answer:

Yeah it's right there from the one next to the exclamation point

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3 years ago

Read 2 more answers

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 $x_f_g$

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

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3 years ago

What does the term mccarthyism describe? To whom does it refer?

kolbaska11 [484]

<span>Mccarthyism describes a political witchhunt, specifically refers to Sen. McCarthy's cold war era congressional commitees whose aim was to investigate and "out" or reveal individuals with communist "sympathies".</span>

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3 years ago

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

Airida [17]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

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3 years ago