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3 years ago

5

Which best describes the relationship between population size, carrying capacity, and limiting factors?

2 answers:

borishaifa [10]3 years ago

7 0

Answer:

is b

Explanation:

Alexandra [31]3 years ago

6 0

When the sources is not enough for a population size that would limit population growth are limiting factors. And these limiting factors prevent the population size to increase. When it stops growing, the ecosystem reaches its carrying capacity. <span>Carrying capacity is the end of the line of a population size that the ecosystem can support over time.</span>

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Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis. The magnitude of q1 is 3 times the magnitu

Alex777 [14]

d = distance between the two point charges = 60 cm = 0.60 m

r = distance of the location of point "a" where the electric field is zero from charge $q_{1}$ between the two charges.

$q_{1}$ = magnitude of charge on one charge

$q_{2}$ = magnitude of charge on other charge

$q_{1}$ = 3 $q_{2}$

$E_{1}$ = Electric field by charge $q_{1}$ at point "a"

$E_{2}$ = Electric field by charge $q_{2}$ at point "a"

Electric field by charge $q_{1}$ at point "a" is given as

$E_{1}$ = k $q_{1}$ /r²

Electric field by charge $q_{2}$ at point "a" is given as

$E_{2}$ = k $q_{2}$ /(d-r)²

For the electric field to be zero at point "a"

$E_{2}$ = $E_{1}$

k $q_{2}$ /(d-r)² = k $q_{1}$ /r²

$q_{2}$ /(d-r)² = 3 $q_{2}$ /r²

1/(0.60 - r)² = 3 /r²

r = 0.38 m

r = 38 cm

8 0

3 years ago

How does the electric field intensity vary with the increase of distance of the point from the centre of a charged conducting sp

Nataly_w [17]

Answer:

(i) Electric field outside the shell:

For point r>R; draw a spherical gaussian surface of radius r.

Using gauss law, ∮E.ds=q0qend

Since E is perpendicular to gaussian surface, angle betwee E is 0.

Also E being constant, can be taken out of integral.

So, E(4πr2)=q0q

So, E=4πε01r2q

4 0

3 years ago

At the instant a traffic light turns green, a car starts with a constant acceleration of 1.3 m/s2. At the same instant a truck,

Bas_tet [7]

Answer:

(a) Distance traveled = 75.3846 m

(b) Velocity of car at that instant will be 14 m/sec

Explanation:

We have given acceleration of the car $a=1.3m/sec^2$

Initial velocity of the cart u = 0 m/sec

(a) According to second equation of motion we know that $s=ut+\frac{1}{2}at^2$

So distance traveled by car $s_c=0\times t+\frac{1}{2}\times 1.3t^2=0.65t^2$

As the truck is moving with constant speed

So distance traveled by truck $s_t=ut=7t$

As the truck overtakes the car

So $s_c=s_t$

$0.65t^2=7t$

$t=10.769sec$

So distance traveled $s_c=s_t=7\times 10.769=75.3846m$

(b) From second equation of motion we know that v = u+at

So v = 0+1.3×10.769 = 14 m /sec

7 0

3 years ago

Atruck with a mass of 5280 Kg has an acceleration of 45 m/s/s what is the force acting on the truck?

horrorfan [7]

117.3 I think that’s it mb if that wrong

8 0

3 years ago

A car ends up at x = -245 m after a displacement of -103 m. what was its initial position?

AysviL [449]

Answer:

-142m

Explanation:

To find the initial position of the car, we can add the displacement.

-245 + 103 = -142

Best of Luck!

5 0

3 years ago