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3 years ago

14

An object accelerates. • 1st law • 2nd law

2 answers:

sergij07 [2.7K]3 years ago

7 0

Answer:

2nd law

Explanation:

hope it helps...

klio [65]3 years ago

3 0

This is a abbreviation of the 2nd Law of Motion

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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

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

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

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

∴

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

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

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

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

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

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

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

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

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

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

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3 years ago

A cube at 101 °C radiates heat at a rate of 67 J/s. If its surface temperature is increased to 166 °C, the rate at which it will

igor_vitrenko [27]

Answer:

The rate at which it will radiate heat is closest to 127.2J.

Explanation:

According to Stefan's law, the heat radiated by a black body can be written as:

$\frac{dH}{dt}$ is directly proprtional to the $T^{4}$ .

where H is the amount of heat radiated and T is the surface temperature .

<u>Example</u> : The heat is transferred by radiation and an example of heat transferred by radiation is the sunlight reaching earth from sun.

<u>Note</u> : Here the temperature is expressed in Kelvin.

Initial rate of heat , $\frac{dH_{1}}{dt}=67J/s$

Final rate of heat , $\frac{dH_{2}}{dt}$ = unknown

Initial temperature , $T_{1}=374K$

Final temperature , $T_{2}=439K$

$\frac{H_1}{T_{1}^4}=\frac{H_2}{T_{2}^4}$

$H_{2}=67\times1.898$

$H_{2}=127.2$ .

The rate at which it will radiate heat is closest to 127.2J.

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