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Sati [7]
3 years ago
6

Index Fossils can help scientists determine the age of a rock layer because they are present in every rock layer since the begin

ning of life on Earth.
Physics
1 answer:
irakobra [83]3 years ago
8 0

Answer:

They were only present during a short time frame, so it helps to narrow down the age of the rock layer.  

Explanation: hope this helps :)

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Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft
Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0
3 years ago
How do the discovery of gravity and the invention of electronic satellites most likely relate to the processes of scientific inv
Leya [2.2K]

Answer:

c

Explanation:

because i said so

5 0
3 years ago
Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
melisa1 [442]

Answer:

120 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 12 m/s²

t = 10 s

Find: v

v = at + v₀

v = (12 m/s²) (10 s) + 0 m/s

v = 120 m/s

6 0
3 years ago
Which nervous system consist of the brain and spine
AURORKA [14]
The CNS (Central Nervous System) consists of the brain and spinal cord.
5 0
3 years ago
#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

V=Ax

where A is the given area, so

4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

8 0
3 years ago
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