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2 years ago

Angles t and v are complementary.angles T has a mesure of (2X+10). Angle v has a measure of 48 what is the value of x

Physics

1 answer:

juin [17]2 years ago

6 0

16 because complementary angles equal up to 90. 2x+58 = 90 x= 16

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D. Budgeting time, avoiding stress, and prioritizing.

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3 years ago

State the reason and tell whether true or false. The SPEED OF LIGHT IS THE SAME IN ALL MEDIA.

Sphinxa [80]

Answer:

<em>The statement is true .</em>

Explanation:

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3 years ago

What is the definition of specific heat capacity

mamaluj [8]

Answer:

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3 years ago

A ruby laser delivers a 16.0-ns pulse of 4.20-MW average power. If the photons have a wavelength of 694.3 nm, how many are conta

stepan [7]

Answer:

The value is $n = 2.347 *10^{17} \ photons$

Explanation:

From the question we are told that

The amount of power delivered is $P = 4.20 \ M W = 4.20 *10^{6} \ W$

The time taken is $t = 16.0ns = 16.0 *10^{-9} \ s$

The wavelength is $\lambda = 694.3 \ nm = 694.3 *10^{-9} \ m$

Generally the energy delivered is mathematically represented as

$E = P * t = \frac{n * h * c }{\lambda }$

Where $h$ is the Planck's constant with value $h = 6.262 *10^{-34} \ J \cdot s$

c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$4.20 *10^{6} * 16*10^{-9}= \frac{n * 6.626 *10^{-34} * 3.0*10^{8} }{694.3 *10^{-9}}$

=> $n = 2.347 *10^{17} \ photons$

4 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