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NISA [10]
3 years ago
13

A student creates a model of a closed ecosystem by filling a glass tank half full with water then adding 10 snails and two small

aquatic plants the next day all the snails are dead what is the most likely cause of their death
Physics
2 answers:
worty [1.4K]3 years ago
8 0

In any closed ecosystem, if we put some snails and some plants in them, the snails would survive for a while but next day they will die, this is because of the reason that the snails can survive for a while in closed ecosystem but the waste products which takes place in the ecosystem due to both the plants as well as the snails affects the snails due to which they are likely to die soon.

GenaCL600 [577]3 years ago
4 0
The reason why the snails are dead because the snails lack oxygen for it is a model of closed ecosystem. Another reason to this is that they lack  abiotic factors that they need in surviving or as they live for the place that they are living lacks a lot and are not enough for them to last or survive. Lastly, the water could have not been a contributor for them to live for they may not be able to live in a water environment and could only dwell in land.
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Answer: 6611.715 joules

Explanation:

Q = MxCxdeltaT = 6959.7 which is 100%

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Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th
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The equilibrium temperature is T13=3.12 ◦C

<u>Explanation:</u>

<u>Given </u>

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

<u>1.When the first two liquids are mixed:</u>

  • mC1(T1-T12)+mC2(T2-T12)=0
  • C1(6-13)=C2(23-13)=0
  • 7C1=10C2
  • C1=1.42C2

<u>2.When the second and third liquids are mixed</u><u>:</u>

  • mC2(T2-T23)+mC3(T3-T23)=0
  • C2(23-26.8)=C3(38-26.8)=0
  • 3.8C2=12.8C3
  • C2=3.36C3

<u>3.When the first and third liquids are mixed:</u>

  • mC1(T1-T13)+mC3(T3-T13)=0
  • C1(6-T13)+C3(38-T13)=0
  • C1=1.42C2  C2=3.36C3
  • C1=1.42C2(3.36C3)
  • C1=4.77C3
  • C1(6-T13)+C3(38-T13)=0
  • 4.77C3(6-T13)+C3(38-T13)=0
  • By solving the equation we get,
  • T13=3.12 ◦C
  • The equilibrium temperature is T13=3.12 ◦C

<u></u>

7 0
3 years ago
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Explanation:

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Most ionic bonds form when electrons from____.
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A bond with elements from B.
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A guitar string is supposed to have a fundamental frequency 256 Hz. It currently has a fundamental frequency 248 Hz when the str
umka2103 [35]

Answer:

The tension to bring the guitar string into tune is 372.95 Hz.

Explanation:

Given;

current frequency, f₁ = 248 Hz

current tension, T₁ = 350 N

fundamental frequency, f₂ = 256

The tension on the string to bring the guitar string into tune is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\f\lambda =  \sqrt{\frac{T}{\mu} } \\\\f^2\lambda^2 = \frac{T}{\mu} \\\\f^2 =  \frac{T}{\mu \lambda^2}\\\\let \ {\mu \lambda^2} = k\\\\f^2 =\frac{T}{k} \\\\k = \frac{T}{f^2} \\\\\frac{T_1}{f_1^2} = \frac{T_2}{f_2^2}\\\\T_2 = \frac{T_1 f_2^2}{f_1^2} \\\\T_2 =  \frac{350 \times  256^2}{248^2} \\\\T_2 = 372.95 \ Hz

Therefore, the tension to bring the guitar string into tune is 372.95 Hz.

3 0
3 years ago
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