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Georgia [21]
3 years ago
13

While driving down his street one evening, Jonah notices that his neighbor has laid out an electric fan for the garbage to pick

up the next morning. Jonah stops and has a look at the fan; it seems to be perfectly fine, so he takes it home with him. The fan would most likely be considered___________
Physics
2 answers:
zimovet [89]3 years ago
7 0

Answer:

<h2>C. abandoned property.</h2>

Explanation:

Options for this question are:

A. lost property.

B. mislaid property.

C. abandoned property.

D.None of these are correct.

Abandoned property refers to an asset that it doesn't have an owner after years of inactivity. In real state, the property would be owned by the state. However, this case is simpler, because the neighbour just threw away the fan, and Jonah picked it up from the garbage. Literally, the neighbour abandoned the article.

On the other hand, other options don't apply, because the fan is not a mislaid property, it's not missing from its owner, it wasn't lost.

Therefore the correct answer is C.

Kryger [21]3 years ago
4 0

Answer: I have no idea either i need help aswell

Explanation:

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When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a
Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

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

Put the value into the formula

\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}

\dfrac{1}{f}=-\dfrac{3}{110}

f=-36.6\ cm

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.    

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}

Here, focal length is positive for converging lens

\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}

\dfrac{1}{v}=\dfrac{367}{20130}

v=54.85\ cm

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0
3 years ago
Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o
valkas [14]
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
y= \frac{m\lambda D}{a}
from the center of the pattern. In the formula, m is the order of the minimum, \lambda the wavelenght, D the distance of the screen from the slit and a the width of the slit.

In our problem, the distance of the first-order band (m=1) is y=0.22 cm. The distance of the screen is D=86 cm while the wavelength is \lambda = 384 nm=384 \cdot 10^{-7}cm. Using these data and re-arranging the formula, we can find a, the width of the slit:
a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm
3 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.
ziro4ka [17]

a) The charge on the outer surface is 1.2\cdot 10^{-12} C

b) The number of ions is 7.5\cdot 10^6

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k = 4.3 is the dielectric constant

\epsilon_0 =8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=5.1\cdot 10^{-9} m^2 is the surface area

d=1.4\cdot 10^{-8} m is the distance between the two plates

Substituting,

C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F

The capacity of the membrane is related to the potential difference between the two surfaces by

C=\frac{Q}{\Delta V}

where here we have

Q = excess charge on one surface

\Delta V = 85.5 mV = 0.0855 V is the potential difference between the two surfaces

Solving for Q, we find

Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C

b)

We said that the net charge on the outer surface is

Q=1.2\cdot 10^{-12} C

The charge of one K+ ions is equal to the electron charge

+e=1.6\cdot 10^{-19} C

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6

Learn more about capacity:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

6 0
3 years ago
Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
son4ous [18]

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

4 0
3 years ago
Is a light on potential or kinetic?
Basile [38]

its the the first one u said

7 0
3 years ago
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