As a liquid is cooled its molecules lose kinetic energy and their motion slows. When they've slowed to where intermolecular attractive forces exceed the collisional forces from random motion, then a phase transition from liquid to solid state takes place and the material freezes
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Answer:
The nearest plant (A) receives 4 times more radiation from the farthest plant
Explanation:
The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of the sphere
I = P / A
P = I A
The area of the sphere is
A = 4π r²
Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets
P = I₁ A₁ = I₂ A₂
I₁ / I₂ = A₂ / A₁
Suppose index 1 corresponds to the nearest planet,
r2 = 2 r₁
I₁ / I₂ = r₁² / r₂²
I₁ / I₂ = r₁² / (2r₁)²
I₁ / I₂ = ¼
4 I₁ = I₂
The nearest plant (A) receives 4 times more radiation from the farthest plant
Answer:
Friction is a force that holds back the movement of a sliding object.
Explanation:
The two types of friction: Static friction and Kinetic friction. Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.
Answer:
10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)
Explanation:
The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.
The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.
We use the first equation of motion for a free-falling body to obtain v as follows;
v = u + gt....................(1)
where g is acceleration due to gravity taken as 9.8m/s/s
It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.
To obtain t, we use the second equation of motion as stated;
![h=ut+gt^2/2.................(2)](https://tex.z-dn.net/?f=h%3Dut%2Bgt%5E2%2F2.................%282%29)
Given; h = 6.10m.
since u = 0 for the vertical motion; equation (2) can be written as follows;
![h=\frac{1}{2}gt^2............(3)](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B1%7D%7B2%7Dgt%5E2............%283%29)
substituting;
![6.1=\frac{1}{2}*9.8*t^2\\6.1=4.9t^2\\hence\\t^2=6.1/4.9\\t^2=1.24\\t=\sqrt{1.24}=1.12s](https://tex.z-dn.net/?f=6.1%3D%5Cfrac%7B1%7D%7B2%7D%2A9.8%2At%5E2%5C%5C6.1%3D4.9t%5E2%5C%5Chence%5C%5Ct%5E2%3D6.1%2F4.9%5C%5Ct%5E2%3D1.24%5C%5Ct%3D%5Csqrt%7B1.24%7D%3D1.12s)
Putting this value of t in equation (1) we obtain the following;
v = 0 + 9.8*1.12
v = 10.93m/s