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4 years ago

14

Recrystallization involves any chemical, physical, or biological changes that take place after sediments are deposited and buria

l and lithification occur when unconsolidated sediments are transformed into sedimentary rocks. True or False?

1 answer:

Taya2010 [7]4 years ago

4 0

Yes it will be true!

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A liquid with a density of 900kg/m 3 is stored in a pressurized, closed storage tank. The tank is cylindrical with a 10m diamete

Orlov [11]

Answer:

18.62 m/s

Explanation:

Given that:

A liquid with a density of 900 kg/m 3 is stored in a pressurized, closed storage tank.

Diameter of the tank = 10 m

The absolute pressure in the tank above the liquid is 200 kPa = 200, 000 Pa

At pressure of 200 kPa ; the final velocity = 0

Atmospheric pressure at 5cm = 101325 Pa

We are to calculate the initial velocity of a fluid jet when a 5cm diameter orifice is opened at point A?

By using Bernoulli's theorem between the shaded portion in the diagram;

we have:

$Pa \ + \ \frac{1}{2} \ \delta \ v^2_1 \ + \ \delta gy_1 = P + \delta gy_2$

$\frac{1}{2} \ \delta \ v^2_1 \ = P + \delta gy_2 - \ \delta gy_1 - Pa$

$\frac{1}{2} \ \delta \ v^2_1 \ = \delta g(y_2 -y_1 )+ ( P - Pa )$

$v_1 \ = \sqrt{ \frac {2 \ \delta g(y_2 -y_1 )+ 2 ( P - Pa )}{\delta}}$

where;

Pa = atmospheric pressure = 101325 Pa

$\delta$ = density of liquid = 900 kg/m³

$v_1$ = initial velocity = ???

g = 9.8 m/s²

$y_1$ = height of the hole from the buttom

$y_2$ = height of the liquid surface from the button

$v_1 \ = \sqrt{ \frac {2*900*9.8(7 - 0.5 )+ 2 ( 200,000 - 101325 )}{900}}$

$v_1 = 18.62 \ m/s$

Thus, the initial velocity of the fluid jet = 18.62 m/s

3 0

3 years ago

How to find the specific heat capacity for a liquid

bazaltina [42]

Answer:

Heat Capacity = E / T.

- Find the difference in temperature for changes of multiple degrees.

- Add the appropriate units to your answer to give it meaning.

- Know that this equation works for cooling objects as well.

- Know that specific heat refers to the energy needed to raise one gram by one degree.

5 0

3 years ago

Problems related to radiation. (a) The temperature of the Sun’s photosphere is 5700 K. Assume it is a blackbody. What is the pea

sineoko [7]

Answer:

a) λ = 5,084 10⁻⁷ m , b) P = 3.63 10²⁶ W , c) P = 5.8 10²⁷ W and d) λ = 2.54 10⁻⁷ m

Explanation:

a) The maximum emission of the sun can be calculated using the Win equation

λ T = 2,898 10⁻³ m.K

λ = 2,898 10⁻³ / T

λ = 2,898 10⁻³ / 5700

λ = 5,084 10⁻⁷ m

λ = 5,084 10⁻⁷ m (1 10⁹ nm / 1m) =

λ = 5,084 10² nm = 508.4 nm

photon in the visible range

b) The emission of the Sun, is described by the Stefan equation

P = σ A e T⁴

Where σ is the Stefan-Boltzmann constant that vslue is 5,670 10-8 W/m²K⁴, A area of the Sun, and e the emissivity that for a perfect black body is 1

In order to use this equation, we must calculate the area of the sun, we consider it a perfect sphere

r = 695,000 km (1000m / 1 km) = 6.95 10⁸ m

Area of a sphere

A = 4π R²

A = 4π (6.95 10⁸8)²

A = 6.07 10¹⁸ m²

P = 5,670 10⁻⁸ 6.07 10¹⁸ 1 5700⁴

P = 3.63 10²⁶ W

c) The new temperature is double the previous one

T = 2 To

Let's substitute in the formula and calculate

P = σ A e (2To)⁴

P = σ A e T⁴ 2⁴

Po = σ A e T4 = 3.63 10 26 W

P = 16 Po

P= 16 (3.63 10²⁶)

P = 5.8 10²⁷ W

d) Let's calculate the explicit value of the temperature and use the Win equation

T = 2 5700

T = 11400K

λ = 2,898 10⁻³ / 11400

λ = 2.54 10⁻⁷ m

λ = 2.54 10²nm = 254 nm

photon in the UV range

5 0

3 years ago

Describe what’s going on in position-time graphs 1 , 3, &5

Verizon [17]

Answer:

Sjsjsnsnusjsbsbwhwjahahjaia

7 0

3 years ago

What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 1

-BARSIC- [3]

Answer:

404K

Explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

$K.E=\frac{3}{2}KT$

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

$T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K$

converting to degree we have $131^{0}C$

4 0

3 years ago