Among the 3 state, solid has a fixed volume while liquid takes the volume of its container. When gas is compressed, it takes the volume of its container likewise.

7 0

3 years ago

Interacting with others while exercising is beneficial to social health because it allows a person to __________.

vlada-n [284]

Feel better and develop communication skills

It's not only the physical well-being that has developed as well as intellectual and emotional aspect of an individual. When having conversation to someone and you are doing something it's also the same of having a multitask work. That all senses response quickly and something is developing in you, at same time you are establishing good rapport towards others.

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6 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

3 years ago

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

Travka [436]

1) In a circular motion, the angular displacement $\theta$ is given by
$\theta = \frac{S}{r}$
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: $S=2600 m$ . Using the information about the radius, $r=0.35 m$ , we find the total angular displacement:
$\theta = \frac{2600 m}{0.35 m} =7428 rad$

2) If we put larger tires, with radius $r=0.60 m$ , the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
$\theta = \frac{2600 m}{0.60 m}=4333 rad$

8 0

4 years ago