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marta [7]
4 years ago
6

Which of the following is an obstacle to creating computer-based models for tracking a hurricane?

Physics
1 answer:
iren [92.7K]4 years ago
5 0

Answer:

4. All of the above I think, not to sure about 1. but the rest are right so im like 90.99999 percent sure good luck

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Which is style of conclusion used in research reports
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A conclusion is, in some ways, like your introduction. You restate your thesis and summarize your main points of evidence for the reader.You can usually do this in one paragraph.
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3 years ago
What kind of medium does ocean waves have
Anika [276]
Ocean waves propagate through the medium called 'sea water'.
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4 years ago
Who developed the steady-state hypothesis for the universe? When was it developed?
Gnoma [55]

Answer:

The steady-state theory was first proposed by Sir James Jeans in the 1920s, but it really got a boost in 1948 when it was reformulated by Fred Hoyle, Thomas Gold, and Hermann Bondi.

Explanation:

True

6 0
3 years ago
Read 2 more answers
You want to close an open door by throwing either a 400-g lump of clay or a 400-g rubber ball toward it. you can throw either ob
gogolik [260]
Refer to the diagram shown below.

Let m =  the mass (g) of the door.
Let v =  the launch velocity
Let u =  the velocity of the door after impact.

Elastic impact (rubber ball):
The rubber ball bounces off the door with presumably elastic impact, which means that both momentum and kinetic energy are conserved.
Conservation of momentum requires that
400v = -400v + mu
Therefore
u=( \frac{800}{m} )v

Inelastic impact (clay):
The clay sticks to the door after impact.
Conservation of momentum requires that
400g = (m+400)u
Therefore
u=( \frac{400}{m+400} )v

When we compare magnitudes of u for the door, we find that
u_{1}=( \frac{400}{m} )(2v), \,\, elastic \\\\ u_{2}=( \frac{400}{m+400} )v , \,\, inelastic
Clearly, the elastic impact creates a greater value of u for the door.

Answer:
The rubber ball creates a larger impulse to the door because the nature of its impact is approximately elastic.


5 0
3 years ago
​A ​very ​long ​insulating ​cylinder ​of ​charge ​of ​radius ​2.50 ​cm ​carries ​a ​uniform ​linear ​density ​of ​15.0 ​nC/m. ​(
Pavlova-9 [17]

Answer:

r2 = 2.401557  cm

distance = 0.10 cm

Explanation:

given data

​radius ​= 2.50 ​cm

​density ​= ​15.0 ​nC/m

voltmeter ​read =  ​175

solution

we know here potential difference that is express as

ΔV = \frac{\lambda }{2\pi \epsilon _o} ln\frac{r2}{r1}     ...........1

so here

ln\frac{r2}{r1} = 2\pi \epsilon _o \times \frac{\triangle V}{\lambda }  

as here \lambda is linear charge density  

\frac{r2}{r1} = e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}  

r2 = r1 × e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}  

r2 = 2.40 × e^{2\pi 8.85\times 10^{-12} \times \frac{175}{15\times 10^{-6} }}  

r2 = 2.401557  cm

and

here distance above surface will be

distance = r1 - r2

distance =  2.50 - 2.40

distance = 0.10 cm

4 0
3 years ago
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