Ask question

Ask question

All categories

2 years ago

12

Which of the following types of valves have a valve stem that moves up and down in a valve guide? Select one: a. Poppet valves b

. Reed valves c. Check valves d. All of the above.

2 answers:

Alja [10]2 years ago

5 0

Answer:c

Explanation:

thats wht it is

puteri [66]2 years ago

4 0

Poppet valves are the one whose valve stem moves up and down in a valve guide.

Answer: A

Explanation

Poppet valve is used for controlling the flow of fluid in the engine.

The valves are present inside a cylindrical structure from where the fluids will get inserted or exhausted.

The timing of input and output as well the quantity of input and output is maintained by Poppet valve.

The valve consists of a large metal plate connected with a long metal rod.

The metal rod is termed as valve stem which is present in the valve guide.

The valve stem will be moving up and down to insert and exhaust the gases respectively in an internal combustion engine.

You might be interested in

HELP ME PLEASE RN

IRISSAK [1]

Answer:

information

Explanation:

see picture

8 0

3 years ago

Air at atmospheric pressure and at 300K flows with a velocity of 1.5m/s over a flat plate. The transition from laminar to turbul

Savatey [412]

Answer:3.47 m

Explanation:

Given

Temperature(T)=300 K

velocity(v)=1.5 m/s

At 300 K

$\mu =1.846 \times 10^{-5} Pa-s$

$\rho =1.77 kg/m^3$

And reynold's number is given by

$Re.=\frac{\rho v\time x}{\mu }$

$5\times 10^5=\frac{1.77\times 1.5\times x}{1.846\times 10^{-5}}$

$x=\frac{5\times 10^5\times 1.846\times 10^{-5}}{1.77\times 1.5}$

x=3.47 m

5 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat

ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n ..........1

and

total cost of capital intensive = $1,244,000 + $92.5 n ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n < $33800 + $143 n

solve we get

n > $\frac{906000}{50.5}$

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > $\frac{1244000}{104.5}$

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0

2 years ago

In a website browser address bar, what does “www” stand for?

Ludmilka [50]

Answer:

www stands for world wide web

Explanation:

It will really help you thank you.

3 0

2 years ago