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densk [106]
3 years ago
6

Seagrasses are flowering plants. Forty-nine species in 12 genera are known. The flowers are simple and open underwater; pollen i

s transported by waves and currents. However, they also reproduce vegetatively via rhizomes and form dense mats or meadows often composed of a single species. Totally immersed in seawater, the long, linear leaves have no stomata. The plants never exchange gases with the atmosphere. They absorb carbon dioxide that is dissolved in the water and use the oxygen they themselves produce during photosynthesis.
Considering the characteristics of seagrass, you will most likely find seagrasses in which ocean zone(s)?
A) the ocean basic and the abyss
B) the continental shelf, slope, and rise
C) the continental shelf and the sunlight zone
D) the continental shelf and slope; the sunlight and twilight zones
Physics
2 answers:
Harrizon [31]3 years ago
8 0

i just took the test and its c

olya-2409 [2.1K]3 years ago
5 0

i just took the test and its c

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A stiff wire 44.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l
Helga [31]

The magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.

Explanation:

It is known that the force acting on a current carrying conductor placed in a magnetic field is

F=BIL sin theta

Here B is the magnetic field, I is the current flowing through the wire and L is the length of the wire which is given as 44 cm.

Since the wire is bended in the middle at right angle so the length of the two sides of the wire will be 22 cm each. Also one part is lying over z axis and another part lies in the plane of xy in the equation of line y = 2x. So the slope of this wire will be

\frac{y}{x} =2

This will be equal to tan θ.

So θ = tan⁻¹ (2) =63.4°

Then, the length of the wire will be written as components of i, j and k.

L = (-22)k+(22) cos ( 63.4) i+(22) sin (63.4)j

L = 0.098 i+0.197 j-0.22k

Then,

F = I (L × B)

F = 20.5 ((0.0985 i + 0.197 j -0.22k) * (0.316 i))

F = 20.5 (\left[\begin{array}{ccc}i&j&k\\0.098&0.197&-0.22\\0.316&0&0\end{array}\right] )

F = 20.5(i(0)-j(0-(-0.22*0.316))+k(0-(0.316*0.197))) = 20.5(-0.069 j-0.062 k)

F = -1.415 j-1.271 k

The magnitude of force on the wire will be

F = \sqrt{(-1.415)^{2}+(-1.27)^{2}  } = \sqrt{3.615}=1.90 N

And the direction can be found by the tan inverse of the ratio of k component to j component of the force.

theta = tan-1(\frac{-1.271}{-1.415})= 41.9 degrees

So the magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.

5 0
4 years ago
(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find
storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ ,  and

$f_{m_2}= 95 \ Hz$

Sampling rate f_s = \ 245 \ Hz

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \  f_{o_2}= f_{m_2} = 95 \ Hz $

when n  = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm  f_s $

$f_{o_1}= \pm 65 \pm 245,\ \  f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \  f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \  f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \  f_{o_2}= 395 \ Hz,585 \ Hz$

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

8 0
3 years ago
Compare and contrast the three <br> classes of levers.
Xelga [282]
<em>Class I .</em> . .
The fulcrum is between the effort and the load.
The mechanical advantage may be any positive number, more or less than ' 1 '.

<em>Class II .</em> . .
The load is between the fulcrum and the effort.
The mechanical advantage is always greater than ' 1 '.

<em>Class III .</em> . .
The effort is between the fulcrum and the load.
The mechanical advantage is always less than ' 1 '.
7 0
3 years ago
a racing car traveling initially at 8.0 m/s accelerates uniformly at 10.0 m/s^2 for 5 seconds. How far does it travel in this ti
Pepsi [2]

The car travels a distance of

(8.0 m/s) (5 s) + 1/2 (10.0 m/s²) (5 s)² = 165 m

7 0
3 years ago
Energy that does not involve the large-scale motion or position of objects in a system is called:
Artemon [7]
This type of energy is called non mechanical energy so the answer is C
8 0
3 years ago
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