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zmey [24]
3 years ago
13

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.
Which plate, the upper or the lower, is positively charged?
Physics
1 answer:
saul85 [17]3 years ago
3 0

Answer:

The upper plate is positive charged.

Explanation:

From the information we have, we know that the plastic bead, carrying a negative charge of  -6 nC, is suspended between the two plates by the force of the electric field between them.

If it is suspended, this means that it is at rest. As we have two forces acting on the bead (gravity and electrostatic force), and gravity is always downward, this means that the electrostatic force must be upwards.

If this force is due to the field, as the force on a negative charge is opposite to the direction of the field, this means that the electric field created by the charge on the plates must be pointed downward.

If it points downward, as the electric field has the same direction that it would have a positive test charge, this means that the charge would go down also, away from the positive charged plate, which must be the upper one.

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5 0
3 years ago
An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o
jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

Substitute the values then we get

-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}

\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}

\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}

v=-\frac{20}{3}=-6.7 cm

Image distance of apple=-6.7 cm

Magnification=m=\frac{v}{u}=\frac{-\frac{20}{3}}{-20}

Magnification of apple=\frac{1}{3}=0.33

Hence, the magnification of apple=0.33

5 0
3 years ago
Briefly describe the formation of the planets from the solar nebula
saul85 [17]
<span>The formation of the Solar System began 4.6 billion years ago with the gravitational collapse of a small part of a giant molecular cloud. Most of the collapsing mass collected in the center, forming the Sun, while the rest flattened into a proto-planetary disk out of which the planets, moons, asteroids, and other small Solar System bodies formed.


Hope it helped
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6 0
3 years ago
A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6
Rina8888 [55]

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force \frac{mv^{2} }{r}. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

7 0
3 years ago
two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y
svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

v_0_x=6380 m/s

v_0_y=6770 m/s

time period of engine running=763 s

Displacement in x=4.50\times 10^6

y=7.27\times 10^6

Using s=ut+\frac{at^2}{2} in x and y direction

x=v_0_x\times t+\frac{at^2}{2}

4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}

4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}

a=-1.23 m/s^2

In y direction

y=v_0_y\times t+\frac{a't^2}{2}

7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}

7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}

a=7.24 m/s^2

x component=-1.23 m/s^2

y component=7.24 m/s^2

3 0
3 years ago
Read 2 more answers
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