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3 years ago

Select all that apply. What factors can limit growth?

2 answers:

Rudiy273 years ago

8 0

A, b and d

Competition for fix resources of course increases the energy expenditure needed to gather those resources and leaves less for a given population. A lower amount of sunlight or water also contributes to decreased energy which lowers growth. Finally, small borders limit growth in that population density can only go so high before the death rate increases beyond the birth rate.

Ugo [173]3 years ago

7 0

Answer:

a b and d plz vote brainliest need 1 more for a higher rank up

Explanation:

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0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

3 0

2 years ago

(1 point) A pilot flies in a straight path for 1 hour and 30 min. She then makes a course correction, heading 10 degrees to the

kati45 [8]

Answer:

The pilot is 2214.22 miles from her starting position

Explanation:

Since the pilot is traveling at a constant speed of 635 mph, the total distance traveled can be easily found as follows:

$1stleg: d_{1} = 1.5*635 = 952.5 mi\\2ndleg: d_{2} = 2*635 = 1270 mi$

There was a 10 degrees deviation, so the angle between the trajectory of both legs is 170 degrees.

The distance we need to find is that from the start of the first leg to the end of the second leg, those three distances form a triangle and since the side we're interested in is opposite to the 170 degrees angle, we can determine its length by the law of cosines:

$d_{3}^{2} =d_{1}^{2} + d_{2}^{2} -2*d_{1}*d_{2}*cos(170)\\d_{3}^{2} =952.5^{2} + 1270^{2} -2*952.5*1270*cos(170)\\d_{3}=2214.22mi$

The pilot is 2214.22 miles from her starting position

8 0

2 years ago

As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force

fgiga [73]

Answer: $107.8\ J$

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

$\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J$