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2 years ago

14

When a craton is exposed at earth's surface, it is called a ________. when a craton is exposed at earth's surface, it is called

a ________. caldera continental shield mid-ocean ridge plate transform fault region?

1 answer:

harina [27]2 years ago

3 0

When a Craton is exposed at earth's surface, it is called a continental shield. Continental shield is any of the large stable areas of low relief in the Earth's crust that are composed of Precambrian Crystalline rocks. Shield areas are regarded as continental nuclei, the observation often being made that most continental shields are bordered by belts of folded rocks of post-Precambrian age.

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A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

2 years ago

Timmy drove 2/5 of a journey at an average speed of 20 mph.

mixer [17]

Answer:

4hr

Explanation:

5 0

2 years ago

Read 2 more answers

If the gauge pressure of a gas is 114 kPa, what is the absolute pressure?

Anastasy [175]

Answer:

D. 214 kPa

Explanation:

The absolute pressure is given by:

$p = p_a + p_g$

where

p is the absolute pressure

$p_a \sim 100 kPa$ is the atmospheric pressure

$p_g$ is the gauge pressure

In this problem, we have

$p_g = 114 kPa$

So, the atmospheric pressure is

$p = 100 kPa + 114 kPa = 214 kPa$

4 0

2 years ago

Read 2 more answers

17. The edges of a rectangular solid have these measures: 1.5 feet by 1½ feet by 3 inches. What is its volume in cubic inches? a

Alisiya [41]

Answer:

c. 972

Explanation:

The volume of a rectangular solid is calculated as the product of its dimensions, that is, its width, its length and its height:

$V=a*b*h$

1 feet is equal to 12 inches, so:

$a=1.5ft*\frac{12in}{1ft}=18in\\b=(1+\frac{1}{2})ft\\b=\frac{3}{2}ft*\frac{12in}{1ft}=18in$

Now, we calculate the volume of the object in cubic inches:

$V=18in*18in*3in\\V=972in^3$

6 0

2 years ago

A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the

kap26 [50]

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

5 0

2 years ago