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3 years ago

What most closley defines population density?

1 answer:

3 0

The carrying capacity of a biological species in an environment is the maximum population size of the species that the environment can sustain indefinitely, given the food, habitat, water, and other necessities available in the environment.

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A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its

Savatey [412]

Answer:

a). $E_{kA}=1.2635 J$

b). $V_{B}=4.535\frac{m}{s}$

c). Δ $E_{t}=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J$

b).

$E_{kB}=\frac{1}{2}*m*v_{B} ^{2}$

$V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}$

c).

net work= EkA+EkB

$E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J$

3 0

3 years ago

A 900 kg car moves around a 500 m radius curve at 25.0 m/s. What is the centripetal force on

Gnesinka [82]

Answer:

9375 N

Explanation:

From the question,

Centripetal force (F) = mv²/r.................. Equation 1

Where m = mass of the car, v = velocity of the car, r = radius of the curve.

Given: m = 900 kg, r = 600 m, v = 25 m/s

Substitute these values into equation 1

F = (900×25²)/600

F = 9375 N.

Hence the centripetal force on the car is 9375 N

3 0

2 years ago

Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially

Gennadij [26K]

The trains take <u>57.4 s</u> to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

$v_A=90 km/h\\ v_B=-80 km/h$

The relative velocity of the train A with respect to B is given by,

$v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h$

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

$t= \frac{d}{v_A_B}$

Substitute 2.71 km for d and 170 km/h for $v_A_B$

$t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h$

Express the time in seconds.

$t=(0.01594h)(3600s/h)=57.39s$

Thus, the trains cross each other in <u>57.4 s</u>.

6 0

3 years ago

Earth's gravitational force just got three times stronger! What happens to your weight?

kicyunya [14]

Your "weight" is the name you give to that gravitational force.
So your question actually says:

"Your weight just got three times stronger !
What happens to your weight ?"

8 0

2 years ago

A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located

zhenek [66]

The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

$\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241 \times 10^{-19 } \ C$

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

6 0

2 years ago