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3 years ago

Please help brainliest ASAP...........

Physics

2 answers:

pantera1 [17]3 years ago

8 0

Answer:

I believe it is gender

Explanation:

ElenaW [278]3 years ago

6 0

I agree I think it’s gender.

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A rocket is being launched straight up. Air resistance is not negligible.

Yanka [14]

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2 years ago

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Which of the following is most closely related to an activated complex?

Anon25 [30]

Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

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3 years ago

45. What is the wovelength of a 30. Hertz periodic

Leviafan [203]

The wavelength is 2m.

Hence, Option c) 2m is the correct answer

Given that;

Frequency; $f = 30Hz$

Speed; $v = 60m/s$

Wavelength; $\lambda =\ ?$

using the expression for the relations between wavelength, frequency and speed of wave:

$\lambda = \frac{v}{f}$

Where $\lambda$ is wavelength, f is frequency and v is speed.

We substitute our given values into the equation

$\lambda = \frac{60m/s}{30Hz}\\\\\lambda = \frac{60m/s}{30s^{-1}}\\\\\lambda = 2m$

The wavelength is 2m.

Hence, Option c) 2m is the correct answer.

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4 0

2 years ago

An object is 1.0 cm tall and its erect image is 5.0 cm tall. what is the exact magnification?

ikadub [295]

The exact magnification of the objects is calculated by dividing the cinema. We calculate it by diving the erect image size by the object size. From the given above, we find the exact magnification by dividing 5.0 cm by 1.0 cm. Thus, the answer would be 5.

7 0

2 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