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2 years ago

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

Physics

1 answer:

blsea [12.9K]2 years ago

7 0

Answer: If you remove all of the hawks, it stabilizes an a new equilibrium.

Explanation: For the ecosystem to stabilize at a new equilibrium, the tertiary consumer (hawks) need to be totally removed from the ecosystem. Since there's no tertiary consumer to checkmate the density of other components of the food chain, it will make form a balanced ecosystem.

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The value of (997)1/3 according to binomial theorem is

postnew [5]

Answer:

9.99

Explanation:

The value of (997)^1/3

(997)^1/3

997 = (1000 - 3)

(1000 - 3)^1/3

Expanding :

[1000(1 - 3/1000)]^1/3

1000^1/3 * (1 - 3/1000)^1/3

Cube root of 1000

10 * (1 - 3/1000 * 1/3)

10 * (1 - 1/1000)

10 * (1 - 0.001)

10(0.999)

= 9.99

Hence, the value of (997)^1/3 according to binomial theorem is 9.99

5 0

1 year ago

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

You are given two rectangular blocks of shiny metal, Block A and Block B, and are asked to determine which one will float in a b

vladimir2022 [97]

Answer:

Explanation:

Volume of block A = 10 x 6 x 1 = 60 cm³

Mass of block A = 630 g

density of mass A = mass / density

= 630 / 60 = 10.5g / cm³

Volume of block B = 5 x 5 x 3 = 75 cm³

Mass of block A = 604 g

density of mass A = mass / density

= 604 / 75 = 8.05 g / cm³

Since density of both A and B are less than that of mercury , both will float in mercury.

7 0

2 years ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

7 0

1 year ago

he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th

AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x $10^{-6}$ T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ $_{0}$ I/2πr

where

B = magnetic field (T)

μ $_{0}$ = 1.26 x $10^{-6}$ (N/ $A^{2}$ )

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.19)

= 4.96 x $10^{-6}$ T

BV = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.41)

= 2.3 x $10^{-6}$ T

hence,

the net magnetic field = BH - BV

= 4.96 x $10^{-6}$ - 2.3 x $10^{-6}$

= 2.66 x $10^{-6}$ T

4 0

1 year ago