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Verizon [17]
3 years ago
10

Which statement is true about the relationship between carrying capacity and population size?

Physics
1 answer:
TiliK225 [7]3 years ago
8 0
Carry capacity determines maximum population size
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What are some examples and non-examples of volume
Jlenok [28]

Answer:

Explanation:

Some correct non-examples are: A glass half-empty; Anything in two dimensions; The amount that covers something.

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3 years ago
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Can anybody help me solve this problem? Thank you so much!
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Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math

at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
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3 years ago
In both a dry cell and a wet cell battery, what causes the production of an electrical current.
koban [17]

Answer:

a. chemical reaction

Explanation:

In both a dry and wet cell battery, electric currents are produced through chemical reactions.

Both cells are together called electrochemical cells.

  • In these cells, chemical reactions produce an electric current.
  • They are devices that converts chemical energy into electrical energy
  • The reactions here are spontaneous redox reactions.
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Radon is the heaviest naturally radioactive ________________ gas. A) noble B) halogen C) group 1 D) diatomic
Sever21 [200]
(A) noble gases / group 18
6 0
3 years ago
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A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi
Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

\Delta x_1=0.50m-0.35m=0.15m

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

k\Delta x_1=m_1g

Solving for the spring constant k, we get:

k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m

Finally, we sum this to the unstretched length to obtain the length of the spring:

x_2=0.35m+0.088m=0.44m

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0
3 years ago
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