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3 years ago

13

Me pueden ayudar es urgente es para mañana son el 25 y 34

1 answer:

RSB [31]3 years ago

7 0

Answer:

no se puede ver las palabras si la en cerraran asi l

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Which of the following statements is true?

andrezito [222]

Answer:

D. On average, a woman's athletic ability peaks at a later age than a man's.

Explanation:

5 0

3 years ago

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.

Novosadov [1.4K]

Answer:

389.78681 K

Explanation:

$P_1$ = Initial pressure = 55.1 mmHg

$P_2$ = Final pressure = 1 atm = 760 mmHg

$T_2$ = Boiling point

$T_1$ = Initial temperature = 35°C

$\Delta H_{vap}$ = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

$ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K$

The normal boiling point of the substance is 389.78681 K

3 0

3 years ago

A chemical is always the same thing as an element .<br><br> True <br> False

olya-2409 [2.1K]

That would be false

Hope this helps :)

3 0

3 years ago

Read 2 more answers

Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle

egoroff_w [7]

Answer:

Explanation:

The magnitude of the electric force on this charged particle A depends upon the following

5. the distance between the point charge Q and the charged particle A

8. the amount of the charge on the point charge Q

9. the magnitude of charge on the charged particle A

7 0

3 years ago

Problem 8 I estimate that the Gauss gun (a solenoid) is wound with 500 turns over a distance of 15cm with an average radius of 1

stellarik [79]

Answer:

Energy stored, U = 66.6 J

Explanation:

It is given that,

Number of turns in the solenoid, n = 500

Radius of solenoid, r = 1.5 cm = 0.015 m

Distance, d = 15 cm = 0.15 m

Let U is the energy stored in the solenoid. Its formula is given by :

$U=\dfrac{1}{2}LI^2$

L is the self inductance of the solenoid

$L=\mu_o N^2A d$

N is the no of turns per unit length

$L=\mu_o (n/d)^2A d$

$L=\dfrac{4\pi\cdot10^{-7}\cdot500^{2}\cdot\pi\cdot\left(0.015\right)^{2}}{0.15}$

L = 0.00148 Henry

$U=\dfrac{1}{2}\times 0.00148\times 300^2$

U = 66.6 J

Out of given options, the correct option for the energy stored in the solenoid is 70 J. So, the correct option is (a) "70 J".

6 0

3 years ago