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What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should ne

Strike441 [17]

Answer:

What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should never cross each other they should never have only one measurement value

5 0

4 years ago

Read 2 more answers

A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe ha

poizon [28]

Answer :

The force needed to move the cylinder is 25.6 N

<h2>Further explanation </h2>

Given that,

Length of the cylinder, l = 0.5 m

Outer diameter of the cylinder, d = 8 cm = 0.08 m

Outer radius of the cylinder, $r=0.04\ m$

Inside diameter of the pipe, d = 8.5 cm = 0.085 m

Inside radius of the pipe, $r=0.0425\ m$

Specific gravity of the oil, $\rho=0.92$

Density of oil, $d=\rho\times \rho_w$

Kinematic viscosity of the oil, $v=5.57\times 10^{-4}\ m^2/s$

Velocity of the cylinder, u = 1 m/s

We need to find the force needed to move the cylinder. Let the force is F.

Specific gravity is defined as the ratio of the density of the substance to the density of water.

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

$v=\dfrac{\mu}{d}$

Where, $d$ = density of oil

And $d=\rho\times \rho_w$ (density of oil = specific gravity × density of water )

$d=0.92\times 10^3\ kg/m^3$

So,

$\mu=v\times d$ ..............(1)

$\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3$

$\mu=0.512\ Pa-s$

The separation between the cylinder and pipe is given by :

$dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m$

$d_p\ and\ d_c$ are diameter of pipe and cylinder respectively.

The mathematical expression for the Newton's law of viscosity can be written as:

$\tau\propto\dfrac{du}{dy}$

$\tau=\mu\times \dfrac{du}{dy}$ ..........(2)

Where

$\tau$ = Shear stress, $\tau=\dfrac{F}{A}$ ............(3)

$\mu$ = viscosity

$\dfrac{du}{dy}$ = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :

$\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}$ ...............(4)

A is the area of the cylinder, $A=2\pi rl$

Equation (4) becomes :

$F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl$ ..............(5)

$A=\pi d\times l$

$A=\pi \times 0.08\ m\times 0.5\ m$

$A=0.125\ m^2$

Now, equation (5) becomes :

$F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl$

$F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2$

F = 25.6 N

<h2>Learn more </h2>

Kinematic viscosity : brainly.com/question/12947932

<h2>Keyword : </h2>

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.

7 0

3 years ago

The acceleration of a particle as it moves along a straight line is given

BARSIC [14]

Answer:

V_t=6 = 32 m/s

Explanation:

The origin is at point 0 with the positive motion to the right

The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:

a=dv/dt

From here we can express dv as:

dv = a dt

Replace a by 2t — 1

dv = (2t — 1) dt

Integrate both sides of equation

$\int\limits^v_a {2t-1} \, dv$

v=t

a=t_0

putting these value in integral

We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values

v — 2 = (t^2 — t) — (0^2 — 0)

From here we can write the expression for v as:

v_t=6=6^2-6+2 (1)

So the velocity at t = 6 s is:

v_t=6 = 32 m/s

V_t=6 = 32 m/s

In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero

To do that, we'll just replace v by 0 in expression (1)

0 = t^2 — t + 2

The roots of the quadratic equation are:

t_1/2=1± √(1^2-4*2*1)/2

Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right

Now that we have all the details, we can correctly draw the path of the particle

We can see from the sketch that the total distance traveled is:

s^T=Δs_0-1

s^T=| s_1 - s_0 |

Replace s_0 by its value

s^T=| s_1 - 1 | (2)

In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time

Since we have already wrote expression for v as function of time (step 2), we'll use expression to get the expression for s

v= ds/dt

Multiply both sides of equation by dt

v dt = ds

Replace v by expression (1)

(t^2 — t + 2) dt = ds

Integrate both sides of equation

$\int\limits^t_b {x} \, dx$

t=s

b=(s=0)

x=(t^2 — t + 2)

dx=ds

putting these value in integral

(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0

Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6

(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1

4 0

4 years ago

The data shows that recently the size of the raccoon population decreased. How will the decrease in the raccoon population affec

LenaWriter [7]

Answer: Increase in the population of lower organisms.

Explanation: Raccoons are mesopredators found in the middle levels of food webs and have critical impacts on the dynamics of many other species. Thus, they greatly contribute to the functioning of the ecosystem.

They are naturally equipped to checkmate the population size of other organism that are below them (i.e., their preys) in the food web. They are; predators, pathogen carriers (such as rabies), and competitors, as they compete with some specialist in the food web.

Hence, the decrease in the population of raccoons will lead to the increase in the population size of lower organisms they prey on in the food web or chain.

3 0

3 years ago

2. Consider Dekker’s algorithm written for an arbitrary number of processes by changing the statement executed when leaving the

neonofarm [45]

Answer:

Algorith does not work.

Explanation:

One of the ways to obtain the Dekker Algorithm is through a change in the declaration, that is, a declaration that can be executed at the exact moment it leaves the critical section. This way it is possible that the statement,

turn = 1-i / * P0 sets turn to 1 and P1 sets turn 0 * /

It can be changed to,

turn = (turn +1) \% n / * n = number or processes * /

The result will allow to define if it works or not, that is, if it is greater than 2 the algorithm will not be able to work.

Given this consideration we can say that,

<em>- The dead lock does not occur, because the mutual is imposed (if a resource unit has been assigned to a process, then no other process can access that resource).</em>

<em>- There is the possibility of starving if the shift is established in a non-contentious process.</em>

Directly it can be concluded that there is a possibility of starvation so the algorithm could not work, despite the fact that mutual exclusion guarantees that a dead block does not occur.

4 0

3 years ago