Ask question

Ask question

All categories

3 years ago

9

In which one of the following four-stroke cylinder arrangements will the intake and exhaust valves open downward into the combus

tion chamber?
A. K-type
B. C-type
C. T-type
D. I-type

1 answer:

Bogdan [553]3 years ago

8 0

The cylinder arrangements that will do the job is: <span>D. I-type

In the i-type cylinder arrangement, </span><span> the intake and exhaust valves are positioned vertically at the top of the cylinder head, and both valves will move downward for the combustion chamber to open.
We could see this type of cylinder arrangement on many automobiles product</span>

You might be interested in

Which country was used for astronaut training due to its moonlike terrain?

serg [7]

Answer:

Húsavík, Iceland, with its volcanic landscape,

6 0

2 years ago

Read 2 more answers

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

3 years ago

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

kaheart [24]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

$\Delta T$ is the temperature of difference that we want to find

$\Delta x=2.5 cm =0.025 m$ represent the thickness of the material

If we solve $\Delta T$ in absolute value from the equation (1) we got:

$\Delta T =\frac{Q \Delta x}{Ak}$

First we convert 3KW to W and we got:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

And we have everything to replace and we got:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

5 0

3 years ago

The sound wave produced from playing the A key one octave above the "Middle C" on the plano produces a sound wave with a frequen

natita [175]

Answer:

<u>Frequency</u>- number of wave cycles that occur in a given amount of time.

<u>Pitch</u>- number of wavelengths in a given amount of time.

<u>Amplitude</u>- fluctuation or displacement of a wave from its mean value. That means how high or low they are away from the center line.

<u>Volume</u>- The perception of loudness from the intensity of a sound wave. The higher the intensity of a sound, the louder it is perceived in our ears, and the higher volume it has.

<u>Wavelength</u>- the distance between the tops of the "waves".

3 0

3 years ago

Read 2 more answers

What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant fig

Igoryamba

Answer:

$A=6.36\times 10^5\ m^2$

Explanation:

It is given that,

Radius of the circle, $r=4.5\times 10^4\ cm=4.5\times 10^2\ m$

The area of the circle is given by :

$A=\pi r^2$

$A=\pi (4.5\times 10^2\ m)^2$

$A=636172.51\ m^2$

or

$A=6.36\times 10^5\ m^2$

As there is no uncertainty given in the radius of the circle. So, the area of the circle is $6.36\times 10^5\ m^2$ . Hence, this is the required solution.

5 0

3 years ago