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Vikentia
16 days ago
6

Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d

año físico a rapideces menores a 8 km/h. Si el material de los parachoques se deforma permanentemente después de una compresión de 1.5 cm, pero permanecen como un resorte elástico hasta ese punto, ¿cuál debe ser la constante efectiva de resorte del material del parachoques, suponiendo que el automóvil tiene una masa de 1050 kg y que se prueba chocándolo contra una pared sólida?
Chemistry
1 answer:
KiRa [2.8K]16 days ago
3 0

Response: k = 23045 N/m

Clarification:

To determine the spring constant, one must consider the maximum elastic potential energy that the spring can withstand. The kinetic energy of the vehicle should equal at minimum the elastic potential energy of the spring when it is fully compressed. Hence, we express it as:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the vehicle = 1050 kg

k: spring constant =?

v: car speed = 8 km/h

x: maximum spring compression = 1.5 cm = 0.015m

You need to resolve equation (1) for k. Beforehand, convert the speed v to meters per second:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant calculates to 23045 N/m

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a movable chamber has a volume of 18.5 L (at temperature of 18.5 C) assuming no gas escapes and the pressure remains constant wh
VMariaS [2860]

Answer:

39 ^\circ C

Explanation:

We have:

V₁ = 18.5 L

T₁ = 18.5° C = 273 + 18.5 = 291.5 K

V₂ = 19.8 L

T₂ =?

Pressure remains constant

Applying the ideal gas law

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{18.5}{291.5}=\dfrac{19.8}{T_2}

T_2 = 312 K

T_2 = 312 -273 =39 ^\circ C

4 0
1 month ago
"solid potassium iodide decomposes into iodine gas and solid potassium. Write a a balanced chemical equation for this reaction"
Alekssandra [2891]


Now, construct a balanced equation:

2KI (s) ----\ \textgreater \ 2K(s) + I_2 (g)

I_2 exists in its gaseous form as a diatomic molecule.
3 0
1 month ago
An experimental drug, D, is known to decompose in the blood stream. Tripling the concentration of the drug increases the decompo
lions [2782]

Answer:

The rate law for the decomposition reaction is:

R=k[D]^2

The unit for the rate constant will be M^{-1}s^{-1}

Explanation:

D\rightarrow Product

The rate law can be expressed as:

R=k[D]^x..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

[D]'=3[D]

R'=9\times R

R'=k[D]'^x...[2]

[1] ÷ [2]

\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}

\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}

9=3^x

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

R=k[D]^2

The unit for the rate constant will be:

k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}

The unit for the rate constant will be M^{-1}s^{-1}.

5 0
1 month ago
A stock solution of Cu2+(aq) was prepared by placing 0.8875 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and dilut
Anarel [2728]

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

  • The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
  • Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}

The molarity of Cu²⁺ is:

\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M

4 0
23 days ago
In KCI how are the valence electrons distributed
eduard [2645]

Answer:

Explanation:

In KCl, the two elements that combine to create KCl are potassium (K) and chlorine (Cl).

Potassium, as a Group 1 element, possesses one valence electron in its outermost shell which it readily donates during bonding. Every element aims to achieve a stable electron configuration, typically with 2 or 8 electrons in its outer shell. Potassium is characterized by its lower electronegativity and higher ionization energy, making it more likely to donate its electron than to accept one. On the other hand, chlorine belongs to Group 17 and has 7 electrons in its outer shell, requiring just one additional electron to complete its octet. Chlorine’s higher electronegativity and lower ionization energy facilitate its tendency to accept an electron rather than donate it.

The bond between potassium and chlorine that results in KCl is termed an electrovalent bond.

Reaction equation:

K + Cl → KCl

3 0
1 month ago
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