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Anuta_ua [19.1K]
3 years ago
11

What level (high, low, or medium) is cumulonimbus at?

Physics
1 answer:
Nataly_w [17]3 years ago
7 0
Cumulonimbus clouds are at a high level in the sky
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#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
ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,
Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

d = \frac{(v_f + v_i)}{2} \times t

here we know that

v_f = 10 m/s

v_i = 0

t = 1.2 s

now we will have

d = \frac{(10 + 0)}{2}\times 1.2

d = 5 \times 1.2

d = 6.0 m

4 0
3 years ago
A car of 1400 kg is subject to multiple forces which produce an acceleration of 3.5 m/s2 directed north. Find the net force.​
Greeley [361]

Answer:

will

Explanation:

3 0
3 years ago
What is acceleration produced by a force of 12 newton exerted on an object of mass 3kg​
jarptica [38.1K]

Answer:

a=F/m

a=12N/3kg  (here newton can be written as kgm/s^2 so kg will be cancelled)

a=4m/s^2

Explanation:

3 0
3 years ago
There are two possibilities for final stage of extremely massive stars. The first is a
makkiz [27]

Answer: Black hole.

Explanation:

As the massive star "compacts" under its own gravity, it triggers a massive supernova, after this point the remains of the star can become a neutron star, which is a very compact star made primarily, as the name says, of neutrons. The other possibility is a black hole, which is a finite region of space wherein it's interior there is a big concentration of mass, which creates a gravitational field strong enough that there is no particle that can escape it.

3 0
3 years ago
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