Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= [tex]\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12[/tex] = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
A battery with an f.e.m. of 12 V and negligible internal resistance is connected to a resistor of 545 How much energy is dissipated by the resistor in 65 s?
Answer:
When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then
R
1
in Figure 1(a) could be the resistance of the screwdriver’s shaft,
R
2
the resistance of its handle,
R
3
the person’s body resistance, and
R
4
the resistance of her shoes.
Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)
Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.
Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).
To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.
According to Ohm’s law, the voltage drop,
V
, across a resistor when a current flows through it is calculated using the equation
V
=
I
R
, where
I
equals the current in amps (A) and
R
is the resistance in ohms
(
Ω
)
. Another way to think of this is that
V
is the voltage necessary to make a current
I
flow through a resistance
R
.
So the voltage drop across
R
1
is
V
1
=
I
R
1
, that across
R
2
is
V
2
=
I
R
2
, and that across
R
3
is
V
3
=
I
R
3
. The sum of these voltages equals the voltage output of the source; that is,
V
=
V
1
+
V
2
+
V
3
.
This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation
P
E
=
q
V
, where
q
is the electric charge and
V
is the voltage. Thus the energy supplied by the source is
q
V
, while that dissipated by the resistors is
q
V
1
+
q
V
2
+
q
V
3
.
Explanation:
Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are
Answer:
Hello your question is incomplete below is the complete question
Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are 295 K when 300 W of power are switched on. 1) Find the plate temperature after 10 minutes.
answer ; 311.36 k
Explanation:
Given data :
sum of mass * specific heat products for a base plate and components ( Mcp )
= 5000 J/K
effective heat transfer coefficient * surface area ( hA ) = 10 W/K
Initial temperature of plate and cooling air temperature( Tc ) = 295 k
power ( Q = W ) = 300 W
a) Determine plate temperature after 10 minutes
10 mins = 600 secs ( t )
heat supplied = change in temp + heat loss
Q * t = mCp ( ΔT ) + hA ( ΔT ) t
300*600 = 5000 * ( T -295 ) + 10 ( T -295 ) * 600
therefore ; T - 295 = 16.363
T = 311.36 K
Discuss why TVET Institutions need advice of the business community in order
to provide good programmes.
Answer:
Without the indispensable advice of the business community, TVET Institutions will be unable to cover the gap in career knowledge required by the business community. To develop workers who possess the knowledge and skills required by today's business entities, there is always the continual need for the educational institutions (gown) to regularly meet the business community (town). This meeting provides the necessary ground for the institutions to develop programs that groom the workforce with skills that are needed in the current workplace. Educational institutions that do not seek this important advice from the business community risk developing workers with outdated skills.
Explanation:
TVET Institutions mean Technical and Vocational Education and Training Institutions. They play an important role in equipping young people to enter the world of work. They also continue to develop programs that will improve the employability of workers throughout their careers. They regularly respond to the changing labor market needs, adopt new training strategies and technologies, and expand the outreach of their training to current workers while grooming the young people for work.
Air at 1 atm and 25◦C blows across a large concrete surface 20 m wide maintained
at 60◦C. The flow velocity is 6 m/s. Calculate the convection heat loss from the
surface.
This is heat transfer convection, mechanical engineering
please solve this question guys I'm gonna really really be appreciate it for you guys
Answer:
Air at 1 atm and 25◦C blows across a large concrete surface 20 m wide maintained
at 60◦C. The flow velocity is 6 m/s. Calculate the convection heat loss from the
surface.
This is heat transfer convection, mechanical engineering
please solve this question guys I'm gonna really really be appreciate it for you guys
An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into the tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45 mm of water below atmospheric. Assume that the air is incompressible, and at 25 C, 100 kPa absolute. Calculate the air speed in the wind-tunnel test section
Answer:
Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s
Explanation:
Given data :
Manometer reading ; p1 - p2 = 45 mm of water
Pressure at section ( I ) p1 = 100 kPa ( abs )
temperature ( T1 ) = 25°C
Pw ( density of water ) = 999 kg/m3
g = 9.81 m/s^2
next we apply Bernoulli equation at section 1 and section 2
p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex] ---------- ( 1 )
considering ideal gas equation
Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )
R ( constant ) = 287 NM/kg.k
T = 25 + 273.15 = 298.15 k
P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2
substitute values into equation ( 2 )
= 100 * 10^3 / (287 * 298.15)
= 1.17 kg/m^3
Also note ; p1 - p2 = PwgΔh ------- ( 3 )
finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )
[tex]\frac{PairV^{2} _{2} }{2}[/tex] = PwgΔh
[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex] = 753.86
[tex]v_{2}[/tex] = 27.5 m/s
Find R subscript C and R subscript B in the following circuit such that BJT would be in the active region with V subscript C E end subscript equals 5 V and I subscript C equals 25 m A V subscript C C end subscript equals 15 space V comma space V subscript D 0 end subscript equals 0.7 space V comma space beta equals 100 comma space V subscript A equals infinity.. Ignore the early effect in biasing calculations.
Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
Think about all the things that you have learned about wildfire in this module.
In light of climate change, explain how this risk map might look different in 100 years.
Answer:
CLIMATE CHANGE HAS inexorably stacked the deck in favor of bigger and more intense fires across the American West over the past few decades, science has incontrovertibly shown. Increasing heat, changing rain and snow patterns, shifts in plant communities, and other climate-related changes have vastly increased the likelihood that fires will start more often and burn more intensely and widely than they have in the past.
Explanation:
The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect the total quantity of bricks used in the wall
True or False
Answer:
false
Explanation:
2. What is the main job of a cylinder head?
OA. Contain the rapid increase in combustion chamber temperature
OB. Contain the rapid increase in combustion chamber pressure
OC. Prevent engine oil from getting past the pistons
OD. Hold the Head Gasket in place
Grade/Exit
Answer:
Explanation:
The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder
the importance of reading a circuit diagram to interpret a wiring diagram?
Answer:
The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.
Explanation:
I need help with part (C). Pleasee help me. It’s due in a few hours.
Answer:
u do the same thing as part B but only add 100 k, I think, cuz I'm still in middle school but I mean if u see it asks u to do the same thing as B but C says that instead, u do it at half pressure and 100 k is higher temp so what its asking is to repeat b but the twist is u do it at half pressure and 100 k is the higher temp
hope this helps :)
How many kg moles of Sodium Sulphate will contain 10 kg of
Sodium?
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
(as per my thinking)
Answer:
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
How do you describe sound? (SELECT ALL THAT APPLY.) PLEASE HELP AND SELECT ALL THAT APPLY PLEASE!! A. Sound waves have to have a medium to travel through. B. The volume of a sound is known as amplitude. C. Loud sounds have high amplitude and vibrate with more energy than soft sounds. D. Sound waves are compression waves that cause energy transfer in air molecules.
Answer:
Sound waves are compression waves that cause energy transfer in air molecules
Sound waves have to have a medium to travel through
Loud sounds have high amplitude and vibrate with more energy than soft sounds
Explanation:
Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.
Sound is an example of a mechanical wave hence it requires a material medium for propagation.
The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.
trevor moves a magnetic toy train away from a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement?
Answer:a
Ieieksdjd snsnsnsnsksks
Forcing a solid piece of heated aluminum through a die forms:
A. a stamped part
B.a cast part
C.an extruded part.
D.a forged part
Answer:
B a cast part
Explanation:
Extrusion is defined as the process of shaping material, such as aluminum, by forcing it to flow through a shaped opening in a die. Extruded material emerges as an elongated piece with the same profile as the die opening.
Forcing a solid piece of heated aluminum through a die forms: a cast part. Hence, option B is correct.
What is cast part?A liquid element is more often filled with concrete that has a hollow chamber in the correct form during the casting manufacturing process, and the item is then let to harden.
A casting, which is the term for the solidified component, is ejected or broken out of the mould to complete the procedure.
Thus, option B is correct.
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According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.
Answer:
high school diploma
Explanation:
According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.
In other words, they do not seek to acquire a post-secondary school education.
Answer:
B
Explanation:
According to edge its answer B
associate's degree or on-the-job experience
got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.
Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
bachelor’s degree
associate's degree or on-the-job experience .......This is the correct answer.
some college, no degree
associate degree
If a poems has a regular rhythm throughout the poem, it has: PLEASE HELP MEH I WILL GIVE YOU BRAINLEIST!!
A. tone
B. imagery
C. irony
D. meter
Answer:
D, meter.
Explanation:
Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.
If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.
What is meter in poem?Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.
The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.
Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.
Therefore, option D is correct.
To learn more about the poem, refer to:
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Thermoplastic parts are
A.commonly used for outer mirror housings.
B. commonly used for grilles
C.formed by stamping a plastic sheet in a mold
D.formed by forcing a molten solution into a mold.
Answer:
c
Explanation:
Thermoplastic parts are formed by forcing a molten solution into a mold.
Thus option D is correct.
Here,
Thermoplastic are plastic parts made from thermoplastic materials. These materials have the ability to be melted and remolded several times without undergoing any chemical change.
The thermoplastic parts are commonly used for different purposes such as in automotive industries, construction, medical, consumer goods, and much more. These parts are easily moldable and can be made into different shapes and sizes. They are also lightweight, strong, and durable, making them ideal for a wide range of applications.
They are commonly used for applications that require high strength and durability, such as in the automotive and aerospace industries.
Therefore option D is correct.
Know more about thermoplastic,
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using credit reduces future income
Answer:
lol
Explanation:r