PLEASE HELP ME!! <br><br> If the refractive index of water is 1.33, then its critical angle is...

Mangrove forests play an important role in _______.

AysviL [449]

D. All of the above

At high tide fish will feed among the mangrove roots - rich fishing ground

The trees trap sediment and soil in the river that would flow out to sea which also helps stop erosion

Wildlife utilise almost every part of the tree, with insects and birds, monkeys and lizards in the branches, shrimps and fish in the roots, and snails and clams in the soil

4 0

2 years ago

Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

GalinKa [24]

Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.

7 0

3 years ago

A meter stick balances horizontally on a knife-edge at the 51 cm mark. With two nickels stacked over the 6.0 cm mark, the stick

Oliga [24]

Answer:

65g

Explanation:

Two main conditions for equilibrium are:

I. The resultant force must be equal to zero. That is, sum of the forces acting in one direction about a point must be equal to the sum of the forces acting in the opposite direction about the same point.

II. The resultant moment must be equal to zero. That is, sum of the moments in one direction about a point must be equal to the sum of the moments in another direction about the same point.

For the above question,

the 51cm mark is the point where the resultant weight of the meter stick lies,

the pivot or point is the 45cm mark where the stick balanced when 2 nickels ( total mass (5.0g x 2) 10g were placed at the 6cm mark.

Using the conversion factor:

1000g(1kg) = 10N, we can convert mass to weight, calculate the weight of the meter stick then reconvert to mass.

That is,

mass of 2 nickels = 10g = 10/1000 = 0.01N.

Moment = Force x distance from line of force to pivot of rotation

Applying the principle of equilibrium,

Moment of left side = Moment of right side

0.01 x (45-6) = W x (51-45)

Where W = weight of the meter stick

W x 6 = 0.01 x 39

W x 6 = 0.39

W = 0.39/6

W= 0.065N

Therefore, mass of meter stick = 0.065 x 1000 = 65g.

4 0

3 years ago

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0

3 years ago

A magnesium oxide component must not fail when a tensile stress of 10.5 MPa is applied. Determine the maximum allowable surface

Aloiza [94]

Answer:

Maximum permitted surface crack length is 1.29 mm

Explanation:

As per the question:

Tensile stress, $\sigma = 10.5\ MPa = 10.5\times 10^{6}\ Pa$

Surface energy of magnesium oxide, SE = $1.0\ J/m^{2}$

Modulus of elasticity of the material, E = 225 GPa = $225\times 10^{9}\ Pa$

Now,

To calculate the maximum allowable surface crack length:

$L = \frac{2E\times SE}{\sigma^{2}\pi }$

$L = \frac{2\times 225\times 10^{9}\times 1.0}{10.5\times 10^{6}\times \pi } = 1.29\times 10^{- 3}\ m = 1.29\ mm$

7 0

4 years ago

PLEASE HELP ME!! If the refractive index of water is 1.33, then its critical angle is...