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2 years ago

A ray of light is incident on a body x what is the reftactive index

Physics

1 answer:

evablogger [386]2 years ago

6 0

Answer:

1.63

Explanation:

If you have the following options:

B. 1.50

C. 1.49

D. 1.33

E. 1.02

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Balboa Park in San Diego has an outdoor organ. When the air temperature increases, the fundamental frequency of one of the organ

musickatia [10]

The fundamental frequency of one of the organ pipes will go up or increase.

When pressured air is forced into an organ pipe, it echoes at a particular pitch, generating the sound of the pipe organ. Each pipe has been adjusted to a particular pitch on the musical scale.

A musical instrument called an outdoor pipe organ is used to perform music. It produces some calming tones and has a really serene sound. The organ pipe produces the sound of the outdoor organ. The wavelength of the sound is also dependent on the length of the pipe. The fundamental frequency of one of the organ pipes will grow as the speed of the sound increases as the ambient air temperature rises.

The correct option is (c).

Learn more about frequency here:

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6 0

1 year ago

A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu

yarga [219]

Answer:

The work done is 5136.88 J.

Explanation:

Given that,

n = 1.90 mol

Temperature = 296 K

If the initial volume is V then the final volume will be V/3.

We need to calculate the work done

Using formula of work done

$W=nRT\ ln(\dfrac{V_{f}}{V_{i}})$

Put the value into the formula

$W=1.90\times8.314\times296\ ln(\dfrac{\dfrac{V}{3}}{V})$

$W=1.90\times8.314\times296\ ln(\dfrac{1}{3})$

$W=−5136.88\ J$

The Work done on the system.

Hence, The work done is 5136.88 J.

5 0

2 years ago

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul

photoshop1234 [79]

1) First of all, let's find the resistance of the wire by using Ohm's law:
$V=IR$
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
$R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega$

2) Now that we have the value of the resistance, we can find the resistivity of the wire $\rho$ by using the following relationship:
$\rho = \frac{RA}{L}$
Where A is the cross-sectional area of the wire and L its length.
We already have its length $L=2.90 m$ , while we need to calculate the area A starting from the radius:
$A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2$

And now we can find the resistivity:
$\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m$

7 0

3 years ago

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$