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2 years ago

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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 typical human contains 5.00 l of blood, and it takes 1.00 min for all of it to pass through the heart when the person is resti

Mrrafil [7]

<span>(a) 0.0676 l (b) 67.6 cc So we've been told that 5.00 L of blood flows through the heart every minute and that the heart beats 74.0 times per minute. So that means that for every beat of the heart, 5.00 L / 74.0 = 0.067567568 L of blood flows through the heart. Rounding to 3 significant figures gives 0.0676 l. Converting from liters to cubic centimeters simply require a multiplication by 1000, so we have 67.6 cc of blood pumped per beat.</span>

5 0

2 years ago

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A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

Lyrx [107]

Answer:

a) 27.3 m/s

b) 3.78 s

c) 10.576 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

a)

$v=u+at\\\Rightarrow v=0+4.2\times 6.5\\\Rightarrow v=27.3\ m/s$

Top speed of the gazelle is 27.3 m/s

b)

$s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0t+\frac{1}{2}\times 4.2\times t^2\\\Rightarrow t=\sqrt{\frac{30\times 2}{4.2}}\\\Rightarrow t=3.78\ s$

The gazelle would take 3.78 seconds to win a 30 m race.

c)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 6.5+\frac{1}{2}\times 4.2\times 6.5^2\\\Rightarrow s=88.725\ m$

The gazelle would go 88.725 m with the acceleration after which there will be no acceleration.

Time = Distance / Speed

$\text{Time}=\frac{200-88.725}{27.3}=4.076\ s$

Total time taken by the gazelle to cover 200 m would be 6.5+4.076 = 10.576 seconds

4 0

2 years ago

. A lightbulb with a resistance of 2.9 ohms is operated using a 1.5-volt battery. At what rate is

Kobotan [32]

Answer:

Explanation:

Given that,

A light bulb has a resistance of 2.9ohms

R = 2.9 ohms

And a battery of 1.5V is applied

V = 1.5 V

We want to find the rate of energy transformed

First we need to know what rate of energy is

Rate of energy implies that we want to find power. Power is the rate at which work is done

P = Workdone / time

Then,

In electronic, the power dissipated by a resistor is given as

P = V² / R

P = 1.5² / 2.9

P = 0.7759 W

P ≈ 0.776 W

So, the rate at which electrical energy transformed in the lightbulb is 0.776 Watts

8 0

2 years ago

Read 2 more answers

Synthetic plastics are made by linking many simple carbon molecules together to form much larger molecules. This process is call

professor190 [17]

Answer:

A. polymerization

Explanation:

Synthetic plastics are made by linking many simple carbon molecules together to form much larger molecules. This process is called polymerization.

Synthetic or artifical giant molecules consists of synthetic polymers such as plastics, elastomers etc. They are made up of simple monomers which links to form the complex and giant structure.

Monomers are the simplest unit of polymers. Polymers have very great sizes. The size mkaes their structure quite complex. This makes the molecules more disposed in a regular pattern with respect to one another.

The complexity of structure and the attendant effects accounts for the properties and uses that makes synthetic molecules very unique. For example, plastics can be extruded as sheets, pipes and or moulded into other objects.

8 0

2 years ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

docker41 [41]

1) At x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) At x = 6.6 cm, $E_y=0$

3) At x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) At x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field is zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

1)

The electric field of an infinite sheet of charge is perpendicular to the sheet:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

The field produced by a thick slab, outside the slab itself, is the same as an infinite sheet.

So, the electric field at x = 6.6 cm (which is on the right of both the sheet and the slab) is the superposition of the fields produced by the sheet and by the slab:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field of the sheet is to the left (negative charge, inward field), while the field of the slab is the right (positive charge, outward field).

So,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

And the negative sign indicates that the direction is to the right.

2)

We note that the field produced both by the sheet and by the slab is perpendicular to the sheet and the slab: so it is directed along the x-direction (no component along the y-direction).

So the total field along the y-direction is zero.

This is a consequence of the fact that both the sheet and the slab are infinite along the y-axis. This means that if we take a random point along the x-axis, the y-component of the field generated by an element of surface dS of the sheet (or the slab), $dE_y$ , is equal and opposite to the y-component of the field generated by an element of surface dS of the sheet located at exactly on the opposite side with respect to the x-axis, $-dE_y$ . Therefore, the net field along the y-direction is always zero.

3)

Here it is similar to part 1), but this time the point is located at

x = 1.45 cm

so between the sheet and the slab. This means that both the fields of the sheet and of the slab are to the left, because the slab is negatively charged (so the field is outward). Therefore, the total field is

$E=E_1-E_2$

Substituting the same expressions of part 1), we find

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative sign indicates that the direction is to the left.

4)

This part is similar to part 2). Since the field is always perpendicular to the slab and the sheet, it has no component along the y-axis, therefore the y-component of the electric field is zero.

5)

Here we note that the slab is conductive: this means that the charges in the slab are free to move.

We note that the net charge on the slab is positive: this means that there is an excess of positive charge overall. Also, since the sheet (on the left of the slab) is negatively charged, the positive charges migrate to the left end of the slab (at a = 2.9 cm) while the negative charges migrate to the right end (at b = 4 cm).

The net charge per unit area of the slab is

$\sigma=+64\mu C/m^2$

And this the average of the surface charge density on both sides of the slab, a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Also, the infinite sheet located at x = 0, which has a negative charge $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the left surface of the slab, so

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Now we have two equations (1) and (2), so we can solve to find the surface charge densities on a and b, and we find:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

6)

Here we want to calculate the value of the x-component of the electric field at

x = 3.34 cm

We notice that this point is located inside the slab, because its edges are at

a = 2.9 cm

b = 4.0 cm

But slab is conducting , and the electric field inside a conductor is always zero (because the charges are in equilibrium): therefore, this means that the x-component of the electric field inside the slab is zero

7)

We calculated the value of the charge per unit area on the surface of the slab at x = a = 2.9 cm in part 5), and it is $\sigma_a = +65.25 \mu C/m^2$

8)

As we said in part 6), the electric field inside a conductor is always zero. Since the slab in this problem is conducting, this means that the electric field inside the slab is zero: therefore, the regions where the field is zero is

2.9 cm < x < 4 cm

So the correct answer is

"none of these region"

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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2 years ago