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2 years ago

6

The process of sediments being compacted and cemented to form sedimentary rocks is called

1 answer:

crimeas [40]2 years ago

8 0

Answer:

Lithification is the answer.

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A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.48° apart. What is the angular

larisa [96]

Answer:

0.38°

Explanation:

$\theta$ = Angle

m = Number

d = Distance

n = Refractive index of liquid = 1.25

a denotes air

l denotes liquid

In the case of double split interferance we have the relation

$m\lambda=dsin\theta$

For air

$m\lambda_a=dsin\theta_a$

For liquid

$m\lambda_l=dsin\theta_l$

Dividing the two equations

$\frac{m\lambda_a}{m\lambda_l}=\frac{dsin\theta_a}{dsin\theta_l}\\\Rightarrow \frac{\lambda_a}{\lambda_l}=\frac{sin\theta_a}{sin\theta_l}$

Wavelength ratio = $n$

$n=\frac{sin\theta_a}{sin\theta_l}\\\Rightarrow \frac{sin0.48}{1.25}=sin\theta_l\\\Rightarrow \theta_l=sin^{-1}\frac{sin0.48}{1.25}\\\Rightarrow \theta_l=0.38^{\circ}$

The angular separation is 0.38°

5 0

3 years ago

Read 2 more answers

a city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in

ivolga24 [154]

We have that for the Question "A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?" it can be said that

OC=2

From the question we are told

A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?

Generally the equation for the Opportunity cost is mathematically given as

$OC=\frac{y_1-y_2}{x_1-x_2}\\\\OC=\frac{2-0}{1-0}\\\\OC=2$

Therefore

From the graph of the question we can ascertain that Opportunity cost

OC=2

For more information on this visit

brainly.com/question/18670421

Graph is attached below

6 0

2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of

lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0

3 years ago

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0

2 years ago