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valentinak56 [21]
3 years ago
13

How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se

a acquire energy directly from the sun. b. Organisms near the ocean’s surface rely on chemosynthesis rather than photosynthesis. c. Organisms in the deep sea do not have direct access to sunlight. d. Organisms in the deep sea have no means to obtain energy unless they travel to the ocean’s surface.
Physics
2 answers:
nadya68 [22]3 years ago
6 0

Answer: Option C

Explanation:

The organism that live in the deep sea obtain energy by hetero tropic mode of nutrition. Even if they will travel to the ocean's surface they will not perform photosynthesis.

So, there is no access of sunlight to the organism that live in deep ocean. The organism have different mode of nutrition.

Some of the organism might perform chemo synthesis deep inside the ocean to obtain energy.

Hence, the correct answer is option C

Montano1993 [528]3 years ago
3 0

that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats


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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th
-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = -13.125 m/s^2

Explanation:

By momentum conservation along x direction we will have

mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2

so we have

v_1 + v_2 = 2v

v_1 + v_2 = 4.68

also by energy conservation

\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J

\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17

(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)

(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83

21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83

2v_2^2 - 9.36v_2 + 2.12 = 0

by solving above equation we will have

v_1 = 4.44 m/s

v_2 = 0.24 m/s

Average acceleration on first part of the chunk is given as

a_1 = \frac{4.44 - 2.34}{0.16}

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = \frac{0.24 - 2.34}{0.16}

a_2 = -13.125 m/s^2

5 0
3 years ago
PLZ ANSWER THE QUESTION ​
adelina 88 [10]

Answer:

a

Explanation:

t

7 0
3 years ago
Hen a gfci receptacle device is installed on a 20-ampere branch circuit (12 awg copper), what is the minimum volume allowance (i
Bezzdna [24]

Answer:

2.25in³

Explanation:

For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³

See attached Table 314.16(B) from NEC 2011

4 0
3 years ago
Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

6 0
3 years ago
A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning
zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

L_{i}=L_{f}

Here,  = the system's angular momentum before the collision

L_{i} = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here,  = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

4 0
1 year ago
Read 2 more answers
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