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lord [1]
3 years ago
10

Geologists use tools other than the law of superposition to help them with relative dating. Which statements describe those tool

s? Check all that apply.
Physics
1 answer:
White raven [17]3 years ago
6 0

Answer: Cross-cutting features are always younger than the surrounding rock.

When material erodes before sediment is deposited on it, a geologic gap results.

Explanation:

The options include:

1. An unconformity is created when lava pours out on Earth’s surface.

2. Faults are the result of volcanic activity.

3. Intrusions and extrusions are sedimentary formations.

4. Cross-cutting features are always younger than the surrounding rock.

5. When material erodes before sediment is deposited on it, a geologic gap results.

The law of superposition simply states that when there is a layers of rocks, we would see that the younger layers will lie and be on top of the layers that are older.

Other tools that can help scientist with relative dating are:

• Cross-cutting features are always younger than the surrounding rock.

• When material erodes before sediment is deposited on it, a geologic gap results.

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4 points
zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

I = k\frac{1}{r^2}

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

I_2 = k\frac{1}{(2r)^2}

Dividing both the equation we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4}  = 25lx

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

I_2 = k\frac{1}{(3r)^2}

Dividing equation 1 and 3 we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9}  = 11.11lx

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0
3 years ago
When you look ahead while driving, it is best to:
vodka [1.7K]
I think its [B]
Personally i would say [B] only because If you are looking beyond the car in front of you..... then what if the car in front of you throws on breaks... you would hit them in the butt because you weren't paying attention to the car.
And majority of the time if your looking in the lanes beside you then you are most likely trying to get in that lane.
4 0
3 years ago
Is India a rich country?
Norma-Jean [14]

Explanation:

India. Total wealth: $8.9 trillion | Wealth per capita: $6,440 | India, which is the fifth-largest economy in the world, is home to 3,57,000 HNWIs and 128 billionaires.

4 0
2 years ago
Read 2 more answers
Which of the following statements is TRUE concerning forces?
Gnesinka [82]
The answer is d I took the test
3 0
3 years ago
A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft
erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)

Now, differentiating both sides with respect to time 't', we get:

2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}

Now, we are given speed as 25 mph. So, \frac{de}{dt}=25\ mph

Also, when e=91\ mi, we can find 'x' using equation (1). This gives,

x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi

Now, plug in the values of 'e' and 'x' and solve for \frac{dx}{dt}. This gives,

\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0
3 years ago
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