Which of the following statements are true with respect to the law of conservation of mass?

What part of a circuit has the main job of conducting the electricity?

Vesnalui [34]

The main job of conducting electricity is the power source.

5 0

3 years ago

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

Aleks04 [339]

I’m pretty sure it’s b

8 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle

Contact [7]

<h2>Answers: </h2>

1) 1.359, 1.403

2) $2.207(10)^{8}m/s$ , $2.138(10)^{8}m/s$

Explanation:

The described situation is known as Refraction.

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (1)

On the other hand we have the Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (2)

Where:

$n_{1}$ is the first medium refractive index . We are told is the air, hence $n_{1}\approx 1$

$n_{2}$ is the second medium refractive index

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=38.1\º$ :

$(1)sin(57.0\º)=n_{2}sin(38.1\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}$

$n_{2}=1.359$ (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=36.7\º$ :

$(1)sin(57.0\º)=n_{2}sin(36.7\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}$

$n_{2}=1.403$ (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

$n_{red}=\frac{c}{v_{red}}$

$v_{red}=\frac{c}{n_{red}}$

Substituting (3) in this equation:

$v_{red}=\frac{3(10)^{8}m/s}{1.359}$

$v_{red}=2.207(10)^{8}m/s$ >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

$n_{violet}=\frac{c}{v_{violet}}$

$v_{violet}=\frac{c}{n_{violet}}$

Substituting (4) in this equation:

$v_{violet}=\frac{3(10)^{8}m/s}{1.403}$

$v_{red}=2.138(10)^{8}m/s$ >>>>Speed of violet light

3 0

3 years ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

3 years ago

Read 2 more answers

A beam of visible light is passed through a plexiglass container filled with water. If a beam of light comes in at a slight angl

pentagon [3]

B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate

8 0

4 years ago

Read 2 more answers