Correct answer is Availability of soil minerals.
There are two types of limiting factors that affect plant population in a ecosystem. They are:
- Biotic
- Abiotic
Biotic factors includes food, diseases etc. Such as Invasive weed species, seed dispersal by wind, disease-causing fungal spores are examples of biotic limiting factor.
Abiotic factors include sunlight, temperature and chemical environment. The availability of soil mineral is an example of abiotic limiting factor. Growth of plant population depends on availability of soil minerals.
Answer:
110.4 v
Explanation:
Find the equivalent resistance of the two parallel circuits
R1 and R2 in parallel = R1 * R2 / (R1+ R2) = <u>1 ohm </u>
similarly R4 and R5 = <u>4.77 ohms</u>
Now you can add the three resistances into one (R3=10)
1 ohm + 10 ohm + 4.77 ohm = 15.77 ohms
Now
V = IR
V = 7 amps * 15.77 ohm = 110.4 v
Answer:
The <u>HIGHER</u> value of coefficient of friction the greater the resistance to sliding
Explanation:
As we know that friction force is given by

so friction force is the product of friction coefficient and normal force.
So here we can say that if friction coefficient is higher then the friction force will be more as it is product of friction coefficient and normal force.
So correct answer will be given as
The <u>HIGHER</u> value of coefficient of friction the greater the resistance to sliding
Answer:
In the Equator:
As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.
In Antarctic:
As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.
Explanation:
Assuming that the stone is thrown vertically... let's say it's a 1 kg stone.It doesn't matter if it's thrown upwards or downwards as (assuming no air friction) it will pass the original throwing point with the same downwards velocity as it had upwards, 3 seconds previously. So it starts with 1/2 m v^2 = 0.5 * 1 * 15^2 = 112.5 J of keThen k.e. gained = gpe lostk.e. gained = m g h = 1 * 10 * 50 = 500 J of Ke gainedso the final (total) ke is 612.5 J which = 1/2 m v^2 = 0.5 v^2 here
so 0.5 v^2 = 612.5so v^2 = 1225so v = 35 m/s