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2 years ago

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Two formats of sources are

1 answer:

Scilla [17]2 years ago

3 0

Answer:

APA and MLA are the two format sources.

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The ________ and ________ are used to resolve vectors into their components.

forsale [732]

The sin function and the cos function are used to resolve vectors into their componets

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3 years ago

In which direction does the electric field point at a position directly east of a

svlad2 [7]

Answer:

<h2>D. East </h2><h3>That's my correct answer</h3>

3 0

2 years ago

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

Nat2105 [25]

Answer:

Explanation:

Given

diameter $d=20 \mu m$

density $\rho =1300 kg/m^3$

frequency $\nu =150 Hz$

Length of silk strand $L=14 cm$

Velocity in the string is as follows

$\nu =\sqrt{\frac{T}{\mu }}$

The expression for Fundamental Frequency

$f=\frac{\nu }{2l}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}$

$f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}$

Squaring

$f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}$

$T=4\rho \cdot A\cdot l^2\cdot f^2$

$T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2$

$T=7.2\times 10^{-4} N$

8 0

3 years ago

One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-

choli [55]

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = $0.12 \times 10^{-2} \ m$

$x = 1.17 \ cm = 1.17 \times 10^{-2}\ m$

m = 149 kg

$\delta = 7.87 \ g/cm^3$

$da = 2.28 \times 10^{-10}\ m$

$F_{net} = F-mg\\ \\0 = F - mg \\ \\ F = mg \\ \\ k_sx = mg \\ \\$

∴

$k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\ k_s = 124803.42 \ N /m$

$N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}$

$N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2$

$N_{chain} = (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2$

$N_{chain} = 2.77 \times 10^{13}$

$N_{bond} = \dfrac{L}{da} \\ \\ = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}$

$\text{Finally; the stiffness of a single interatomic spring is:}$

$k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s$

$k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)$

$\mathbf{k_{si} =45.46 \ N/m}$

4 0

2 years ago

A 30 cm scale has one end broken. The mark at the broken end is 2.6 cm. How would you use it to measure the length of your penci

lidiya [134]

Answer:

the mark of the broken end is 2.6 cm so, we use the scale from the next full mark i.e. 3cm

Explanation:

<em>we </em><em>now </em><em>measure</em><em> </em><em>the </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>by </em><em>keeping </em><em>the </em><em>3</em><em> </em><em>c</em><em>m</em><em> </em><em>mark </em><em>of </em><em>the </em><em>scale</em><em> </em><em>at </em><em>it's</em><em> </em><em>left </em><em>end.</em>

<em>The </em><em>3</em><em> </em><em>cm </em><em>value </em><em>is </em><em>then </em><em>subtracted</em><em> </em><em>from </em><em>the </em><em>scale</em><em> </em><em>reading</em><em> </em><em>at </em><em>the </em><em>right</em><em> </em><em>side </em><em>end </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>to </em><em>obtain </em><em>the </em><em>correct</em><em> </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil.</em><em> </em><em>✏️</em>

<em>(</em><em>i </em><em>i </em><em>)</em><em> </em>place the scale in the contact with object along it's length

(2) Your eyes must be exactly in front of the point where the measurements to be taken.

Hope_it_helps_mga_ka_joiners_mwehehe

6 0

2 years ago