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Greeley [361]
2 years ago
11

A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg

e of -6.0 nC, is suspended between the two plates by the force of the electric field between them.
Required:
a. Which plate, the upper or the lower, is positively charged?
b. What is the charges on positvie plate?
Physics
1 answer:
marissa [1.9K]2 years ago
3 0

Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

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Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.
Mars2501 [29]

Answer:

hello your question has some missing parts below is the complete question

and the missing diagram

The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.

answer : 1316.2 hertz

Explanation:

The frequency that produce the maximum sound intensity can be calculated using the relation below

dsin ∅ = n <em>A</em>

where <em>A = </em>dsin ∅ / n  when n = 1 . d = 0.800

<em>A</em> = 0.800 * ( 1 / 3.162 )

<em>A</em> = 0.253 m

speed of sound = 333 m/s

frequency = speed /<em> A</em>

<em>=   </em>333 / 0.253 =  1316.2 hertz

7 0
2 years ago
A projectile was launched horizontally with a velocity of 468 m/s, 1.86 m above the ground. Calculate how long it would take for
elena-14-01-66 [18.8K]

Answer:

  0.6 seconds

Explanation:

The time to fall from height h is ...

  t = √(2h/g)

  t = √(2(1.86 m)/(9.8 m/s^2)) ≈ √0.3796 s ≈ 0.616 s

It would take about 0.6 seconds for the projectile to hit the ground.

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2 years ago
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Bats use a process called echolocation to find their food. This involves giving out sound waves that hit possible prey or food.
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The type of waves used by bats are sound waves. Most of the species use their larynx to produce ultrasound waves in the frequency range of 20 to 200 kilohertz.
These sound waves are echoed, reflected, by surroundings, in this case food or prey. These reflections are received by the specialized receptor cells in the ears of bats. The reflections are analyzed by the brain to make an image.
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2 years ago
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dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
2 years ago
A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved​ along, a marble was fir
bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.  

So with y(1) = v - 4.9 = 0 we have  

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0
2 years ago
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