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zzz [600]
3 years ago
8

Problem 18 if anyone could help that would be amazing!

Physics
1 answer:
svlad2 [7]3 years ago
3 0

i THINK question 18 is (b) and 19 are (A)

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I think this one's B. energy and work are both measured in joules.
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3 years ago
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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
3 years ago
PLEASE HELP ASAP!!
kolezko [41]

A. Wedge

and

B.Screw

A screw is like an inclined plane wrapped around a pole or rod.

and a wedge looks kind of looks like this / and is inclined.

Hope this helps!

Please mark me brainliest! Thanks!

5 0
3 years ago
Physical activity is the most important thing you can do to improve and maintain health-related physical fitness. Which list con
katovenus [111]

Answer:

I guess it's d because age environment heredity Nd maturation r more obvious factors if we compre with only physical fitness

8 0
3 years ago
The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the
Studentka2010 [4]

Answer:

V(t)= 240V* H(t-5)

Explanation:

The heaviside function is defined as:

H(t) =1 \quad t\geq 0\\H(t) =0 \quad t

so we see that the Heaviside function "switches on" whent=0, and remains switched on when t>0

If we want our heaviside function to switch on when t=5, we need the argument to the heaviside function to be 0 when t=5

Thus we define a function f:

f(t) = H(t-5)

The -5 term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (H(t-5) =1 when t\geq 5, so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

V(t)= 240V* H(t-5)

I have made a sketch for you, and added it as attachment.  

5 0
3 years ago
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