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3 years ago

14

In the diagram, above, marker F is pointing to a __________, which are formed when meanders wear away at a narrow point and a po

ol of water forms.

2 answers:

castortr0y [4]3 years ago

4 0

The answer is <span>oxbow lake</span>

zubka84 [21]3 years ago

3 0

B. oxbow lake, gl on the program to you smart environmental scientists.

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Friction exists only when two objects rub against each other.<br> A.True<br> B.False

Rus_ich [418]

A. True friction is two objects touching each other

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3 years ago

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Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

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3 years ago

Mechanical (sound) waves to Earth from satellites. How is this possible? aves are unable to travel through a vacuum, such as thr

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Sound waves are not able to travel in a vacuum, sound requires a medium such as air or a solid for example. Satellites are indeed in space, but radio waves are not sound waves they are a form of light. Light can travel in a vacuum so the messages that satellites beam back to Earth are light waves not sound waves, that is why it is possible.

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3 years ago

The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

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Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

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3 years ago

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How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$

Generally, the equation for Rate of flow of Liquid is mathematically given as

$\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}$

$$

Where dP is pressure difference r is the radius

$\mu$ is the viscosity of water

L is the length of the pipe

$Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}$

$Q=5.2 \mathrm{~mL} / \mathrm{s}$

In $30s the quantity that flows out of the tube

$&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}$

In conclusion, the quantity that flows out of the tube

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

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2 years ago