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3 years ago

Following could help prevent flood waters from running off too quickly causing further soil erosion

1 answer:

8 0

Growing grass plants and trees is the easiest and best way to prevent flood waters from running off too quickly causing further soil erosion. <span />

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Sound is an example of ?

wlad13 [49]

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

8 0

3 years ago

Read 2 more answers

The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0

3 years ago

A 2000 kg roller coaster is at the top of a loop with a radius of 24 m. If its speed is 18 m/s at this point, what force does it

levacccp [35]

Answer:

$46620\ \text{N}$

Explanation:

m = Mass of roller coaster = 2000 kg

r = Radius of loop = 24 m

v = Velocity of roller coaster = 18 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Normal force at the point will be

$N-mg=\dfrac{mv^2}{r}\\\Rightarrow N=\dfrac{mv^2}{r}+mg\\\Rightarrow N=\dfrac{2000\times 18^2}{24}+2000\times 9.81\\\Rightarrow N=46620\ \text{N}$

The force exerted on the track is $46620\ \text{N}$ .

8 0

3 years ago

Monochromatic light falls on two very narrow slits 0.048 mm apart. successive fringes on a screen 5.00 m away are 6.5 cm apart n

atroni [7]

In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
$y= \frac{m \lambda D}{d}$
where D=5.00 m is the distance of the screen from the slits, and
$d=0.048 mm=0.048 \cdot 10^{-3}m$ is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
$\lambda = \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}= 6.24 \cdot 10^{-7}m$

And from the relationship between frequency and wavelength, $c=\lambda f$ , we can find the frequency of the light:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz$

4 0

3 years ago

A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D

RideAnS [48]

Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

v² = u² + 2a s

vf² = vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M vi² + FD) / 1/2 M vi²

RK = 1 + FD / 2 M vi²

4 0

3 years ago