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Savatey [412]
3 years ago
12

A positively charged light metal ball is suspended between two oppositely charged metal plates on an insulating thread as shown

below. After being charged once, the plates are disconnected from the battery. Describe the behavior of the ball. Please use 3 content related sentences. (
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

The positive ball would go first to the negatively charged plate.

Explanation:

After which, it would hold a more negative charge. Due to the negative charge, it would travel towards the positive plate. Thereby, it would transfer negative electrons to the positive plate once more. In doing so, it would transfer positive protons to the negative plate. After which, it would hold more  negative electrons and be drawn towards the positive plate once more. The Process would continue until the once-positive and once-negative became neutral ( and were discharged.) Additionally, the ball hanging on the insulator thread would also be neutral and discharged.

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I NEED HELP ON 2 QUESTIONS PLEASEEE
tino4ka555 [31]

Answer:

2) c) give-way vessel

3) a) With one short blast

Explanation:

2) A vessel that is required to take early substantial action to ensure avoiding  collision called Give way vessel

In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel

Therefore, the correct option is c) give-way vessel

3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast

Therefore, the correct option is a) With one short blast.

4 0
3 years ago
When one observes the phase of matter of water, one observes a property of matter.
FromTheMoon [43]

Water is a very unique substance because it can exist in all three phases of matter (solid, liquid, gas) within the normal temperature ranges found on Earth. When one observes the phase of matter of water, one observes a property of matter.

8 0
3 years ago
What acceleration will a force of 20 newtons cause if applied to a go-kart with a mass of 20 kilograms?
iren2701 [21]

Answer:

1m/s is the acceleration used. C

Explanation:

please mark brainliest

3 0
3 years ago
The wave length of violet light rounded to the nearest nanometer is a __ nm
nataly862011 [7]
Wavelength= speed / frequency 

so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)

in scientific notation= 4.13×10^(-7) 

in nanometer = 413 nm
6 0
3 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
4 years ago
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