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4 years ago

How do foraminifera found in rock layers above the k-t boundary compare to those in rock layers below?

1 answer:

8 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below are the choices that can be found elsewhere:

Foraminifera above the boundary are larger and more diverse than those below.
Foraminifera above the boundary are larger and less diverse than those below.
Foraminifera above the boundary are smaller and more diverse than those below.
Foraminifera above the boundary are smaller and less diverse than those below

The answer is Foraminifera above the boundary are smaller and less diverse than those below

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Can Jet Fuel burn through Steel Pipes?

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Burning jet fuel is not hot enough to fully melt steel. Burning jet fuel, however, is more than hot enough to weaken steel.

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3 years ago

A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it

soldi70 [24.7K]

Answer:

The position function is $s_{t}=2t^3+5t^2-5t+9$ .

Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

We know that,

The acceleration is the rate of change of velocity of the particle.

$a = \dfrac{dv}{dt}$

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

The acceleration is

$a_{t} = 12t+10$

$\dfrac{dv}{dt} = 12t+10$

$a_{t}=dv=(12t+10)dt$

On integration both side

$\int{dv}=\int{(12t+10)}dt$

$v_{t}=6t^2+10t+C$

At t = 0

$v_{0}=0+0+C$

$C=-5$

Now, On integration again both side

$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

$s_{t}=2t^{3}+5t^2-5t+C$

At t = 0

$s_{0}=0+0+0+C$

$C=9$

$s_{t}=2t^3+5t^2-5t+9$

Hence, The position function is $s_{t}=2t^3+5t^2-5t+9$ .

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3 years ago

16. Two capacitors have an equivalent

Gennadij [26K]

Answer:

C1 + C2 = 30 parallel connection

C1 * C2 / (C1 + C2) = 7.2 series connection

C1 * C2 = 7.2 * (C1 + C2) = 216

C2 + 216 / C2 = 30 using first equation

C2^2 + 216 = 30 C2

C2^2 - 30 C2 + 216 = 0

C2 = 12 or 18 solving the quadratic

Then C1 = 18 or 12

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3 years ago

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V: there would be no seasons and humanity would suffer

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3 years ago

A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and

ad-work [718]

We can calculate the length of each spring by using the relationship:
$F=kx$
where
F is the force applied to the spring
k is the spring constant
x is the length of the spring (measured with respect to its rest position)

Re-arranging the equation, we have
$x= \frac{F}{k}$

The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
$x_A= \frac{F}{k_A}= \frac{60 N}{4 N/m}=15 m$
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
$x_B= \frac{F}{k}= \frac{60 N}{5 N/m}=12 m$

Therefore, the correct answer is
D.Spring A is 3 m longer than spring B because 15 – 12 = 3.

5 0

3 years ago

Read 2 more answers