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iragen
6 days ago
5

A major disturbance that caused the ecosystem to stabilize at a new equilibrium?

Physics
1 answer:
serg [3.2K]6 days ago
7 0
The complete removal of all hawks allows for stabilization at a new equilibrium.
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A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat
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The focal length of the lens while in water is noted to be 150 cm, whereas in air, it measures 60 cm. To derive these values, the formula incorporates the variations in the refractive index of glass compared to that of the surrounding medium.
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7 days ago
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(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
serg [3228]

Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

Thank you for your attention; I trust this is beneficial to you.

4 0
1 month ago
A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.
serg [3228]

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where

k represents Coulomb's constant

q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C denotes the charge magnitude on the first cork

q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N

B) Attractive

<pas per="" coulomb="" law="" the="" orientation="" of="" electric="" force="" between="" two="" charged="" entities="" relies="" on="" their="" charge="" signs.=""><pmore specifically="">

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) 2.69\cdot 10^{13}

The total charge of the negative cork is

q_2 = -4.3 \cdot 10^{-6}C

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

e=-1.6\cdot 10^{-19}C

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

q_2 = Ne

<pafter solving="" for="" n="" we="" can="" determine="" the="" count="" of="" excess="" electrons:="">

N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}

D) 3.75\cdot 10^{13}

The overall charge on the positive cork is

q_1 = +6.0\cdot 10^{-6}C

<pthe charge="" of="" a="" single="" electron="" is="" known="" to="" be="">

e=-1.6\cdot 10^{-19}C

<pthe total="" charge="" of="" the="" positive="" cork="" results="" from="" n="" excess="" electrons="" which="" can="" be="" depicted="" as="">

q_1 = -Ne

<pby calculating="" for="" n="" we="" derive="" the="" number="" of="" electrons="" cork="" has="" lost:="">

N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}

</pby></pthe></pthe></pafter></pthe></pwe></pthus></pin></pmore></pas>
6 0
27 days ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
Keith_Richards [2897]

Answer:

The flux across the cube's surface is 2.314\ Nm^{2}/C.

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = 8.0\times 10^{- 2}\ m.

Volume charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}.

Now,

To find the electric flux:

\phi = \frac{q}{\epsilon_{o}}

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of vacuum.

The volume charge density for this scenario is described by:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}

Cube volume, V = a^{3}.

Thus,

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}.

The total charge can be derived from equation (2):

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}.

q = 2.048\times 10^{-11}\ F = 20.48\ pF.

Now, insert the value of 'q' into equation (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C.

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1 month ago
A steady circular __________ light means drivers must stop at a marked stop line.
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B. red................
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