Divide by 6 and divide the caw of squaw squaw
Answer:
Consider a system consisting of 4 resources of same type that are share by 3 processes each of which needs at most two resources.Now we will show that the system is deadlock free.
If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done.
Consider a system with m resources of same type being shared by n processes. Resources can be requested and released by processes only on at a time. The system is deadlock free if and only if The sum of all max needs is < m+n .
We can understand the notion of a deadlock from the following simple real-life example.To be able to write a letter one needs a letter pad and a pen. Suppose there in one letterpad and one pen on a table with two persons seated around the table. We shall identify these two persons as Mr. A and Ms. B. Both Mr. A and Ms. B are desirous of writing a letter. So both try to acquire the resources they need. Suppose Mr. A was able to get the letter pad. In the meantime, Ms. B was able to grab the pen. Note that each of them has one of the two resources they need to proceed to write a letter. If they hold on to the resource they possess and await the release of the resource by the other, then neither of them can proceed. They are deadlocked. We can transcribe this example for processes seeking resources to proceed with their execution. Consider an example in which process P1 needs three resources r1 ; r2, and r3 before it can make any further progress. Similarly, process P2 needs two resources r2 and r3 Also, let us assume that these resources are such that once granted, the permission to use is not withdrawn till the processes release these resources. The processes proceed to acquire these resources. Suppose process P1 gets resources r1 and r3 and process P2 is able to get resource r2 only. Now we have a situation in which process P1 is waiting for process P2 to release r2 before it can proceed. Similarly, process P2is waiting for process P1 to release resource r3 before it can proceed. Clearly, this situation can be recognized as a deadlock condition as neither process P1 nor process P2 can make progress. Formally, a deadlock is a condition that may involve two or more processes in a state such that each is waiting for release of a resource which is currently held by some other process.
Answer:
The correct answers are:
a. % w = 33.3%
b. mass of water = 45g
Explanation:
First, let us define the parameters in the question:
void ratio e = 
= 
Specific gravity 
= 
% Saturation S = 
× 
= 
× 
water content w = 
= 
a) To calculate the lower and upper limits of water content:
when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.
when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.
Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.
To get the relationship between water content and saturation, we will manipulate the equations above;
w = 
Recall; mass = Density × volume
w = 
From eqn. (2) = 
∴ 
putting eqn. (6) into (5)
w = 
Again, from eqn (1)
substituting into eqn. (7)
∴ 
With eqn. (7), we can calculate
upper limit of water content
when S = 100% = 1
Given, 
∴
∴ %w = 33.3%
Lower limit of water content
when S = 1% = 0.01
∴ % w = 0.33%
b) Calculating mass of water in 100 cm³ sample of soil (
)
Given, 
, S = 50% = 0.5
%S = 
× = 
×
0.50 = 
mass of water = 
Answer:
The oil reduced friction between the moving parts of the skateboard. ( A )
Explanation:
The oil reduced Friction between the moving parts of the skateboard and this is because the reduction in friction between moving parts causes an increase in speed.
Remy will oil the moving parts that connects the wheels of the skateboard to the Board, because this is where the most friction is found. the friction between the wheels and the ground cannot be affected by oiling the wheels
Answer:
48.61
Explanation:
See attached diagram.
The level rise in the tube is l sin α.
The level drop in the cylinder (let's call it y) is:
π/4 D² y = π/4 d² l
D² y = d² l
y = l (d/D)²
The elevation difference is the sum:
l sin α + l (d/D)²
l (sin α + (d/D)²)
From Bernoulli's principle:
P = ρgl (sin α + (d/D)²)
Divide both sides by density of water (ρw) and gravity:
P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)
h = S l (sin α + (d/D)²)
If we disregard the level change in the cylinder:
h = S l (sin α)
We want the percent error between these two expressions for h to be 0.1% when α = 25°.
[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001
[ S l (d/D)² ] / [ S l (sin α + (d/D)²)] = 0.001
(d/D)² / (sin α + (d/D)²) = 0.001
(d/D)² = 0.001 (sin α + (d/D)²)
(d/D)² = 0.001 sin α + 0.001 (d/D)²
0.999 (d/D)² = 0.001 sin α
d/D = √(0.001 sin α / 0.999)
When α = 25°:
d/D ≈ 0.02057
D/d ≈ 48.61
