Answer:
The first step in turning a rock into a sediment is Compaction
.
Explanation:
Lithification is a process of changing rock into sediments. There are two steps for a rock to lithify. These steps are as follows
- The first step of lithification is compaction where sediments are erosed together by weight of the it. Thus, the upper layers of sediments causes compaction of lower layers.
- The next process of lithification is cementation. In this fluids fill the space between the loose particles.
Hence, the first step for turning rock into sediments is compaction.
Answer:
Explanation:
Speed = distance / time
Velocity = displacement / time
So ,
Speed = 50 km / 0.5 hr = 100 km/h
Velocity = 40 km / 0.5hr = 80 km/h
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
Answer:
The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s
= 16.125 Pa
Explanation:
The Bernoulli's equation is essentially a law of conservation of energy.
It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.
For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.
We also assume that the initial velocity of wind is 0 m/s.
This calculation is presented in the attached images to this solution.
Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.
The density is obtained to be 1.29 kg/m³.
Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.
We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.
Hope this Helps!!!
Answer:
51.82
Explanation:
First of all, let's convert both vectors to cartesian coordinates:
Va = 36 < 53° = (36*cos(53), 36*sin(53))
Va = (21.67, 28.75)
Vb = 47 < 157° = (47*cos(157), 47*sin(157))
Vb = (-43.26, 18.36)
The sum of both vectors will be:
Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:
