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2 years ago

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Scientists are able to use body waves to determine what makes up the different layers of the Earth's interior. What characterist

ics to the body waves have allow scientists to do this?
Rayleigh waves travel through liquids and solids while Love waves only travel through solids.
P-waves travel through liquids and solid while S-waves only travel through solids.
Love waves travel through liquids and solids while Rayleigh waves only travel through solids.
S-waves travel through liquids and solids while P-waves only travel through solids.

1 answer:

k0ka [10]2 years ago

5 0

Answer:

P-waves travel through liquids and solid while S-waves only travel through solids.

Explanation:

Scientists are able to use the fact that P-waves travel through both solids and liquids and waves travel through only solids to determine what makes the different layers of the Earth.

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A spotlight on the ground is shining on a wall 24m away. If a woman 2m tall walks from the spotlight toward the building at a sp

Lubov Fominskaja [6]

Answer:

$\dfrac{dy}{dt}=-0.059\ m/s$

Explanation:

It is given that,

Distance between the spotlight and the wall, y = 24 m

Height of the woman, h = 2 m

The woman walks toward the building at the rate of 0.6 m/s, $\dfrac{dx}{dt}=0.6\ m/s$

In the attached figure, triangle ABC and MNC are similar. So,

$\dfrac{2}{y}=\dfrac{x}{24}$ ............(1)

$y=\dfrac{48}{x}$

When she is 2 meters from the building. So x = 24-2 = 22 m

$y=\dfrac{48}{22}=2.18\ m$

Differentiating equation (1) i.e.

$xy=48$

$x.\dfrac{dy}{dt}+y.\dfrac{dx}{dt}=0$

$22.\dfrac{dy}{dt}+2.18\times 0.6=0$

$\dfrac{dy}{dt}=-0.059\ m/s$

So, her shadow is decreasing at the rate of 0.059 m/s. Hence, this is the required solution.

7 0

4 years ago

element x has two isotopes: x-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and x-29 has an

meriva

Answer:

27.336

Explanation:

The atomic mass is the weighted average of the atomic masses of the isotopes. Simply multiply each isotopes's atomic mass by its relative abundance, then sum the results.

m = 0.8233 × 26.975 + 0.1767 × 29.018

m = 27.336

4 0

3 years ago

A 20 kg block rests on a rough horizontal table. A rope is attatched to the block and is pulled with a force of 80 N to the left

Pani-rosa [81]

Given :

Mass of block , M = 20 kg .

Force applied , F = 80 N .

Acceleration of block , $a=2.5\ m/s^2$ .

To Find :

The coefficient is Kinetic force friction between the block and the table .

Solution :

We know , Force equation on block is given by :

$F_{net }=F-\mu_k mg \\\\ma = F-\mu_k mg \\\\20\times 2.5 = 80 -\mu_k \times 20 \times 10\\\\\mu_k\times 200=30\\\\\mu_k=\dfrac{30}{200}\\\\\ mu_k=0.15$

Therefore , coefficient is Kinetic force friction between the block and the table is 0.15 .

Hence , this is the required solution .

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4 years ago

sasho [114]

As a liquid is heated, its vapor pressure increases until the vapor pressure equals the pressure of the gas above it. ... In order to form vapor, the molecules of the liquid must overcome the forces of attraction between them. The temperature of a boiling liquid remains constant, even when more heat is added.

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3 years ago

Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

koban [17]

Answer:

A

$N = 1340.86 \ slits / cm$

B

$\theta = 15.7^o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 650 \ nm = 650 *10^{-9} \ m$

The angle of first bright fringe is $\theta = 5^o$

The order of the fringe considered is n =1

Generally the condition for constructive interference is

$dsin (\theta ) = n * \lambda$

=> $d = \frac{1 * 650 *10^{-9 }}{ sin(5)}$

=> $d = 7.458 *10^{-6} \ m$

Converting to cm

$d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm$

Generally the number of grating pre centimeter is mathematically represented as

$N = \frac{1}{d}$

=> $N = \frac{1}{0.0007458}$

=> $N = 1340.86 \ slits / cm$

Considering question B

From the question we are told that

The first wavelength is $\lambda_1 = 650 \ nm = 650 *10^{-9} \ m$

The second wavelength is $\lambda_2 = 429 \ m = 420 *10^{-9 } \ m$

The order of the fringe is $n = 2$

The grating is $N = 5000 \ slits / cm$

Generally the slit width is mathematically represented as

$d = \frac{1}{N }$

=> $d = \frac{1}{ 5000 }$

=> $d = 0.0002 \ c m = 2.0 *10^{-6} \ m$

Generally the condition for constructive interference for the first ray is mathematically represented as

$d sin(\theta_1) = n * \lambda_1$

=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

4 0

3 years ago