<span>Prior to the Cambrian, organisms with hard parts (eg. shells and bones) weren't present for most the time, and it's generally such hard parts that get to fossilize. There's also been more time for Precambrian deposits to get destroyed by natural processes, </span>
<span>The supply of Precambrian fossils has actually greatly improved over the last few decades. The so called Ediacaran faunas have been found at many parts of the world, and would be worth looking into.hope it helps</span>
Answer:
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms
2 ).
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
23 =20 3 N
While the uertical component of the force acting in upward direction=Fsin30=40× 21
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN = 528.64
=5.73m/s 2
Answer:
n_cladding = 1.4764
Explanation:
We are told that θ_max = 5 °
Thus;
θ_max + θ_c = 90°
θ_c = 90° - θ_max
θ_c = 90° - 5°
θ_c = 85°
Now, critical angle is given by;
θ_c = sin^(-1) (n_cladding/n_core)
sin θ_c = (n_cladding/n_core)
n_cladding = (n_core) × sin θ_c
Plugging in the relevant values, we have;
n_cladding = 1.482 × sin 85
n_cladding = 1.4764
To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be



Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:



Therefore the magnetic dipole moment is 
Explanation:
Given that,
Weight of water = 25 kg
Temperature = 23°C
Weight of mass = 32 kg
Distance = 5 m
(a). We need to calculate the amount of work done on the water
Using formula of work done



The amount of work done on the water is 1568 J.
(b). We need to calculate the internal-energy change of the water
Using formula of internal energy
The change in internal energy of the water equal to the amount of the work done on the water.


The change in internal energy is 1568 J.
(c). We need to calculate the final temperature of the water
Using formula of the change internal energy





The final temperature of the water is 23.01°C.
(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.
The amount of heat is 1568 J.
Hence, This is the required solution.