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Salsk061 [2.6K]

2 years ago

14

2. Particle motion in a longitudinal wave is ______.

1 answer:

kotykmax [81]2 years ago

6 0

The answer would be A!

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. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the

vichka [17]

Transverse wave as the wave is going up and down no compressions

3 0

2 years ago

A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

2 years ago

A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t

Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature $T_1=323K$

Volume $V_1=V\ and\ V_2=\frac{V}{2}$

As there is no heat transfer so this is an adiabatic process

For and adiabatic process $TV^{\gamma -1}=constant$

Here $\gamma =1.4$

So $T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}$

$T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1$

$T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K$

4 0

2 years ago

8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti

Marrrta [24]

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

$\phi_1 = B.A$

$\phi_1 = (0.60)(\pi (0.05)^2)$

$\phi_1 = 4.7 \times 10^{-3} Wb$

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

$EMF = \frac{\Delta \phi}{\Delta t}$

so we will have

$EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}$

$EMF = 9.4 V$

7 0

2 years ago

Hydrogen (H-1), deuterium (H-2), and tritium (H-3) are the three isotopes of hydrogen. What are the values of a, b, and c in the

asambeis [7]

I believe it is a=1 b=2 c=3

plato students

6 0

2 years ago

Read 2 more answers