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dangina [55]
3 years ago
11

Oceanic water particles mainly move in circles; is this movement greater on the ocean's surface or below the surface? Explain yo

ur reasoning.
Physics
1 answer:
goblinko [34]3 years ago
5 0
I think that the oceanic water particles mainly move in circles greater in the oceans surface because of how big the waves can be and how wind and air impact the motion. The water particles move more on the surface because of the other factors that impact it such as people, wind, air, etc...
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In order for a ball to move upward, the initial velocity of the ball must be greater than _____.
ch4aika [34]

Answer:

The answer is zero please Give me Brainly

Explanation:

5 0
2 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
3 years ago
A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.
Anestetic [448]

Answer:

\omega=0.37 [rad/s]  

Explanation:

We can use the conservation of the angular momentum.

L=mvR

I\omega=mvR

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

(I_{proffesor - stool}+mR^{2})\omega=mvR

Now, we just need to solve it for ω.

\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}

\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}      

\omega=0.37 [rad/s]  

I hope it helps you!

5 0
4 years ago
PLEASE HELP ME I HAVE BEEN DOING THIS FOR HOURS!! How do some carnivorous plants avoid beginning the digestion process from an a
konstantin123 [22]

Answer:

Once a carnivorous plant has procured an item for dinner, it has to have some way to turn it into fertilizer. What carnivorous plants do is very similar to what humans do with their dinner after they have eaten it. Most carnivorous plants have glands that secrete acids and enzymes to dissolve proteins and other compounds. The plants may also enlist other organisms to help with digestion. The plants then absorb the nutrients made available from the prey.

Drosera releases digestive juices through the glands at the tip of its tentacles and absorbs the nutrients through the tentacles, leaf surface, and sessile glands. In order to do this it bends its tentacles and rolls or bends the leaf to get as many tentacles as possible into contact with the prey for digestion and to make as much leaf surface available for absorption. Its relative Drosophyllum has differently structured, non moving tentacles and doesn't use them directly for digestion. Instead it has specialized glands on the surface of the leaf that release the digestive enzymes (see Carniv. Pl. Newslett. 11(3):66-73 ( PDF ) for drawings and discussion).

The sealed trap of Dionaea does digestion in a way similar to the leaf surface digestion carnivores—upon capture of a prey, digestive enzymes in mucous are released. The advantage of the sealed trap of Dionaea is rain won't wash away the nutrients as digestion proceeds.

The sealed trap carnivores Aldrovanda and Utricularia already have water in their traps so they only need to release enzymes. Utricularia appears to release the enzymes continuously into its traps.

The other carnivorous plants use either a mixed mode of digestive enzymes and partner organisms (Genlisea, Sarracenia, most Nepenthes, Cephalotus, some Heliamphora, Roridula) or other organisms exclusively for digestion (most Heliamphora, some Nepenthes, Darlingtonia). Part of the reason for partnering with other organisms is that the plants actually have little choice in the matter. This could also be a factor for the leaf surface and sealed trap digesters as well. The prey will have gut flora that are quite capable of digesting their host when it dies. In addition, insect larvae, frog tadpoles, and predacious protozoans will or will attempt to take up residence in water-filled traps. The plant releasing digestive enzymes and acids into the traps will help tip the nutrition balance to themselves, but there are limits.

Explanation:

6 0
3 years ago
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