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3 years ago

Why are diamonds so good at cutting things answers?

2 answers:

4 0

Diamonds are so good at cutting things, because they are very hard minerals. They can cut other minerals, but other minerals can’t cut them. Out of all the different kinds of minerals on the Mohs’ hardness scale, diamonds rank right at the top, at the hardness of 10. That means, that they are the hardest out of all of the other minerals. The higher the number, the harder it is to be cut.

balu736 [363]3 years ago

3 0

Because they have sharp edges

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What three things are needed for a mechanical wave

dedylja [7]

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Featured snippet from the web

Mechanical waves move energy through a medium by vibrating particles. Mechanical waves can't move energy through a vacuum because there is no matter inside of a vacuum. The three types of mechanical waves are transverse waves, surface waves, and longitudinal waves.

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3 years ago

a spaceship is traveling at a speed of 15000 km/s from planet b toward planet a the spaceship sends out a signal with a waveleng

Bumek [7]

Answer: 4nmeter

Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is

V which is the speed of space craft 15000km/s = 15000000m/s

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3 years ago

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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

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(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

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$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

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3 years ago

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Katarina [22]

Answer:

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Explanation:

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2 years ago

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g100num [7]

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