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3 years ago

What are the pioneer species that can break down rock during primary succession ?

Physics

1 answer:

mojhsa [17]3 years ago

7 0

C is the answer! hope it helps

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Remoras are small fish that attach themselves to the sides of sharks. They get protection and scraps of food from the sharks. Th

mixas84 [53]

The answer is A. Commensalism. <span>Commensalism, in ecology, is a class of relationships between two organisms where one organism benefits from the other without affecting it. This was the briefest answer from Wiki. Hope it helps, and sorry I'm late.</span>

5 0

3 years ago

What is equilibrium????

IrinaK [193]

Answer:

Explanation:

An equilibrium is a state in which opposing forces or influences are banned.

An example of equilibrium is in economics when supply and demand are equal. An example of equilibrium is when you are calm and steady. An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.

7 0

3 years ago

Read 2 more answers

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .

We know that the surface area and volume of the sphere is given by:

$A=4 \pi r^{2}\\V=\frac{4}{3} \pi r^{3}$

Therefore, the ratio between the surface area and the volume for the sphere will be:

$\frac{A}{V}=\frac{4 \pi r^{2}\\}{\frac{4}{3} \pi r^{3}}=\frac{3}{r}$

Equating the volume to the constant c, we will find the value of $r$ .

$V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }$

Substituting the value of r in the ration between surface area and volume, we get:

$\frac{A}{V}=\frac{3}{ (\frac{3c}{4\pi} )^{\frac{1}{3} }}$

Calculating the constants, we get:

$\frac{4.83598}{c^{\frac{1}{3} } }$

Hence, the ration between surface area and volume is $\frac{4.83598}{c^{\frac{1}{3} } }$

To learn more about surface area and volume of sphere, refer to:

brainly.com/question/4387241

#SPJ4

3 0

1 year ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago