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Alex777 [14]
3 years ago
6

You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,

you then drop five stones and record the time for each to impact: 2.19s, 2.30s, 2.26s, 2.29s, and 2.27s. What is the best estimate of the depth of the well? Explain your reasoning.
Physics
1 answer:
telo118 [61]3 years ago
6 0

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

t_{1}=2.19\ sec

t_{2}=2.30\ sec

t_{3}=2.26\ sec

t_{4}=2.29\ sec

t_{5}=2.27\ sec

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

t_{1}=2.19\approx 2.2\ sec

t_{2}=2.30\approx 2.3\ sec

t_{3}=2.26\approx 2.3\ sec

t_{4}=2.29\approx 2.3\ sec

t_{5}=2.27\approx 2.3\ sec

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

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A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

fe =  focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

m_{e} = 1+\frac{25}{1}

m_{e} =26

m = m_{o}* m_{e}

m = 14*26 = 364

4 0
3 years ago
Waves combine to produce a smaller or zero-amplitude wave in a process called
vampirchik [111]
ANSWER: Destructive Interference

-This is the exact definition of the question you have provided, look this term up if you do not believe lol.

Hope this Helps!
6 0
3 years ago
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A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge
Kisachek [45]

Answer:

\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

5 0
3 years ago
A steel pot has a bottom with a cross sectional area of .1m2 and has a thickness of 1cm. The pot is filled with boiling water an
jenyasd209 [6]

Answer:

840000 J/min

Explanation:

Area = A = 0.1 m²

Bottom of pot temperature = 200 °C

Thermal conductivity = k = 14 J/sm°C

Thickness = L = 1 cm = 0.01 m

Temperature of boiling water = 100 °C

From the law of heat conduction

Q = kAΔT/L

⇒ Q = 14×0.1×(200-100)/0.01

⇒ Q = 14000 J/s

Converting to J/minute

Q = 14000×60 = 840000 J/min

∴ Heat being conducted through the pot is 840000 J/min

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How do water and gravity work together to erode soil, sediment, and rock?
TiliK225 [7]

Gravity can cause water to move down and carry sediment and rocks along the bottom. if the water slows down, it drops the rocks and if it floods a field then the grass will also filter smaller particles out leaving fine sediment behind. In addition to it the fast flow of water can cause erosion, over time it can cause valleys as well.

7 0
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