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sertanlavr [38]

4 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]4 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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B. If you ranked yourself from 6 – 10, describe why it is important to you to be respectful in sports and in other activities, a

kaheart [24]

Answer:

Respect is having a regard for other people and their lives; it is showing those around us compassion and empathy. Children who show respect will find they are successful in all aspects of life. The sports environment is a great place to grow and establish respect.

3 0

2 years ago

The cable holding a 2125 kg elevator has a maximum strength of 21,750 N. What is the maximum upward acceleration the cable can g

vivado [14]

Answer:

10.23m/s^2

Explanation:

GIven data

mass of elevator = 2125 kg

Force= 21,750 N

Required

The maximum acceleration upward

F= ma

a= F/m

a=21,750/2125

a= 10.23m/s^2

Hence the acceleration is 10.23m/s^2

4 0

3 years ago

A long solenoid that has 810 turns uniformly distributed over a length of 0.380 m produces a magnetic field of magnitude 1.00 10

IRINA_888 [86]

Answer:

0.037 A

Explanation:

Magnetic field = B = 1.00 e-4 T

Length = L = 0.380 m

Number of turns = 810

B = μ₀ N I / L

⇒ Current = I = B L / μ₀ N = ( 1 e-4) ( 0.380) / (4π × 10⁻⁷)(810)

= 0.037 A = 37.3 mA

5 0

3 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

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3 years ago

¿Por qué los mensajes llegan en forma inmediata de un celular a otro?

KATRIN_1 [288]

Porque el internet ayuda a la rapidez del mensaje.

7 0

3 years ago