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2 years ago

How does the width of the floodplain change over time?

1 answer:

8 0

Floodplains are landscapes shaped by running water. As streams and their larger forms, rivers, flow across the surface of land, they transport eroded rock and other material. (Erosion is the gradual wearing away of Earth surfaces through the action of wind and water.) At points along that journey, when their flow slows, the material they carry is dropped to create what are termed depositional landforms. Among these landforms are deltas and floodplains.

Read more: http://www.scienceclarified.com/landforms/Faults-to-Mountains/Floodplain.html#ixzz7BNHuUb00

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How much power does it take to lift 30.0 kg 100 m high in 5.00 s?

zheka24 [161]

._.Answer:.

ijji._. ji

Explanation:.

5 0

3 years ago

Suppose there is a bright fringe at P Would this bright fringe move closer to O, move further away, or be unchanged, t 1.5 marks

fiasKO [112]

Answer:

Explanation:

The distance of a fringe from centre is proportional to wavelength of light

and inversely proportional to separation of slits. The expression for distance x is given by

x = nλ D / d

where λ is wave length , D is screen distance and d is slit separation.

So first option only is correct because

1 ) the wavelength of blue light is less than that of red

2) Intensity of light does not affect distance of fringe from the centre.

Diffraction symbolises bending of light around sharp edges like slits or boundaries of opaque objects etc.Due to this reason , we do not observe sharp boundary of shadow of an object. Instead around the boundary of shadow, we observe bands of bright and dark color which are also called fringes.

The phenomena of diffraction is explained by wave theory of light.

3 0

3 years ago

in mammal the weight of the heart approximately 0.5% total body weight write linear model that gives the heart weight in terms o

maks197457 [2]

Let the total weight of the body be = x

Let the weight of the heart of the mammals be = y

Now the given percentage of 0.5 %, we will convert it into decimal form,

so this is equal to 0.005 in decimal form,

so the linear equation that gives the heart weight in terms of body weight will be,

y = 0.005x

7 0

4 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago

The red giant Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that lar

Nat2105 [25]

Answer:

Explanation:

a )

Radius of the sun = .69645 x 10⁹ m .

600 times = 600 x .69645 x 10⁹ m

= 4.1787 x 10¹¹ m .

surface area A = 4π (4.1787 x 10¹¹)²

= 219.317 x 10²²

energy radiated E = σ A Τ⁴

= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴

= 100695 x 10²⁶ J

To know the wavelength of photon emitted

$\lambda_mT= b$

$\lambda_m= \frac{b}{T}$

= 2.89777 x 10⁻³ / 3000

= 966 nm

= 1275 /966 eV

1.32 x 1.6 x 10⁻¹⁹ J

= 2.112 x 10⁻¹⁹ J

No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹

= 47677.5 x 10⁴⁵

= .476 x 10⁵⁰ .

b )

energy radiated by our sun per second

E₂ = σ A 5800⁴

energy radiated by Betelgeuse per second

E₁ = σ x 600²A x 3000⁴

E₁ / E₂ = σ x 600²A x 3000⁴ / σ A 5800⁴

= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸

= 25.76 x 10⁸ x 10⁻⁵

= 25760 times .

4 0

4 years ago