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3 years ago

Which of the follow statements best describes the relationship between carrying capacity and population size?

Physics

1 answer:

mihalych1998 [28]3 years ago

5 0

<span>B. Carrying capacity determines maximum population size.</span>

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A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved

oksian1 [2.3K]

Answer:

Explanation:

The volume of contaminated water

= cross sectional area x height of water level

3.14 x 9 x 9 x 7.5 ft³

= 1907.55 ft³

mass = density x volume

= 1907.55 x 63.5 lbs

m = 121129.425 lbs

This mass has to be raised to the height of 8 ft before evacuation .

There is a rise of centre of mass of

8 - 7.5/2 ft

h = 4.25 ft

Energy required

= mgh

= 121129.425 x 32 x 4.25

= 16473601.8 unit.

3 0

3 years ago

Read 2 more answers

A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a

dybincka [34]

Here we can say that rate of flow must be constant

so here we will have

$A_1v_1 = 18 A_2v_2$

now we know that

$A_1 = 1 cm^2$

$A_2 = 0.4 cm^2$

now from above equation

$1 cm^2 v_1 = 18(0.400 cm^2)v_2$

$\frac{v_2}{v_1} = \frac{1}{18\times 0.4}$

$\frac{v_2}{v_1} = 0.14$

so velocity will reduce by factor 0.14

3 0

3 years ago

In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f

Inga [223]

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

Let $m_1, m_2 \text{ and}\ m_3$ be the masses of the three fragments.

Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

$mv=m_1v_1 +m_2v_2+m_3v_3$

$10 \times 2 \hat i=3 \times 12 \hat{j} + 3(v_{2x} \hat{i}+v_{2y} \hat{j})-4 \times 9 \hat{j}$

So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

6 0

3 years ago

The beginning development of a star is marked by a supernova explosion, with the gases present in the nebula being forced

GenaCL600 [577]

The beginning development of a star is marked by a supernova explosion, with the gases present in the nebula being forced to scatter. As the star shrinks, radiation of the surface increases and create pressure on the outside shell to push it away and forming a planetary nebula or white dwarf.

4 0

3 years ago