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3 years ago

Floodplains are most often found for rivers that exist on

Physics

2 answers:

Schach [20]3 years ago

4 0

Answer:

areas that get high rainfall... im pretty sure

Explanation:

Kazeer [188]3 years ago

3 0

I am pretty sure that floodplains are most often found for rivers that exist on 
hilly areas at the base of mountains. In order to give yoy ans example which will make sure that this answer is quite a suitable one, nice example of floodplains
is The Virgin River at the upper end of Zion Canyon. It will definitely help you! Regards.

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Tara prepared a report to show how the amplitude of waves affects the energy of waves. Is her graphical representation correct?

Gelneren [198K]

Answer:

It is not correct because the amplitude of the waves can be bigger than others and the graph can be going up and down

Explanation: I got the question right

6 0

2 years ago

Read 2 more answers

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

2 years ago

A certain heat engine takes in 300 J of energy from a hot source and then transfers 200 J of that energy to a colder object. Wha

Greeley [361]

Answer:

The efficiency is 0.33, or 33%.

Explanation:

From the thermodynamics equations, we know that the formula for the efficiency of a heat engine is:

$\eta=1-\frac{Q_2}{Q_1}$

Where η is the efficiency of the engine, Q_1 is the heat energy taken from the hot source and Q_2 is the heat energy given to the cold object. So, plugging the given values in the formula, we obtain:

$\eta=1-\frac{200J}{300J}=0.33$