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sertanlavr [38]

3 years ago

11

The standard molar heat of vaporization for water is 40.79 kj/mol. How much energy would be required to vaporize 4.72 x 10^10 mo

lecules of water

1 answer:

stellarik [79]3 years ago

7 0

(5.81 x 10^10 molecules) / (6.022 x 10^23 molecules/mol) x (40.79 kj/mol) = 3.94 x 10^-12 kj

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What is the period? Blank seconds.

FromTheMoon [43]

Answer:

Period refers to the time for something to happen and is measured in seconds/cycle

8 0

3 years ago

A circuit has a 10 Ω resistor connected to a 1.5 V dry cell. What is the current that can flow in the circuit?

ludmilkaskok [199]

Here is the highly detailed, arcane, complex, technical form of Ohm's Law that is needed in order to answer this question ===> I = V / R .

Current = (voltage) / (resistance)

Current = (1.5 V) / (10 Ω)

<em>Current = 0.15 Ampere</em>

8 0

2 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago

An amateur astronomer looks at the moon through a telescope with a 15-cm-diameter objective. What is the minimum separation betw

Lana71 [14]

Answer:

y = 128.0 km

Explanation:

The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains

the diffraction equation for the first minimum is

a sin θ = λ

In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.

θ = 1.22 λ / d

d = 15 cm

to find the distance we can use trigonometry

tan θ = y / L

tan θ = sin θ / cos θ = θ

substituting

y / L = λ / d

y = L λ /d

let's calculate

y = 384 10⁸ 500 10⁻⁹ / 0.15

y = 1.28 10⁵ m

Let's reduce to km

y = 1.28 10⁵ m (1km / 10³ m)

y = 128.0 km

the correct answer is 120 km away

5 0

3 years ago

A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien

Neporo4naja [7]

Answer:

The magnetic field is $B = 8.20 *10^{-3} \ T$

Explanation:

From the question we are told that

The mass of the metal rod is $m = 0.12 \ kg$

The current on the rod is $I = 4.1 \ A$

The distance of separation(equivalent to length of the rod ) is $L = 6.3 \ m$

The coefficient of kinetic friction is $\mu_k = 0.18$

The kinetic frictional force is $F_k = 0.212 \ N$

The constant speed is $v = 5.1 \ m/s$

Generally the magnetic force on the rod is mathematically represented as

$F = B * I * L$

For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

$F_ k = B* I * L$

=> $B = \frac{F_k}{L * I }$

=> $B = \frac{0.212}{ 6.3 * 4.1 }$

=> $B = 8.20 *10^{-3} \ T$

7 0

3 years ago