Answer:
Maybe when there is a fire there can be fire drones that can take it out. and it can also resuce people who are stuck there.
Explanation:
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:
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:
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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Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.
What is forging?Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.
Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.
The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.
Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.
Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.
Thus, the answer is cold forging.
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A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa.
Answer:
a) for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b) for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Explanation:
Given that;
A₂ = 0.001 m²
P₁ = 1 MPa
T₁ = 360 K
k = 1.4
P₂ = 500 Kpa
(1000/500)^(1.4-1 / 1.4) = 360 /T₂
2^(0.4/1.4) = 360/T₂
1.219 = 360 / T₂
T₂ = 360 / 1.219
T₂ = 295.32 K
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
we substitute
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
1.005 × 360 = 1.005 × 295.32 + v₂²/2000
v₂ = 360.56 m/s²
p₂v₂ = mRT₂
500 × (0.001 × 360.56) = m × 0.287 × 295.32
m = 2.127 kg/s
so Mach Number = V₂ / Vc
Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s
So Mach Number = V₂ / Vc = 360.56 / 344.47 = 1.046
Therefore for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b)
AT P₂ = 784 kPa
(1000/784)^(1.4-1 / 1.4) = 360/T₂
T₂ = 335.82 K
now
V₂²/2000 = 1.005( 360 - 335.82)
V₂ = 220.45 m/s
P₂V₂ = mRT₂
784 × (0.001 × 220.45) = m( 0.287) ( 335.82)
172.83 = 96.38 m
m = 172.83 / 96.38
m = 1.793 kg/s
just like in a)
Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s
Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6
Therefore for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Following are the
Given:
[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]
To find:
Flow rate of mass, and Mach number
Solution:
For point a)
Using formula:
[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]
[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]
Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]
[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]
[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]
For point b)
[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]
now
[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]
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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.
determine the values of the viscous damping coefficient c for which the system has a damping ratio of 0.5 and 1.5
Answer:
7.47 lb. s/ft
22.42 lb. s/ft
Explanation:
Without mincing words let us dive straight into the solution to the above problem.
STEP ONE: calculate or determine the frequency.
The frequency can be calculated by making use of the formula given below;
frequency, w = √k/m. Thus, frequency = √ 12 × 15/ [40/32.2] = 12 rad/s.
STEP TWO: Determine or calculate for the viscous damping coefficient, c for damping ratio of 0.5 and 1.5 respectively.
The viscous damping coefficient, c for 0.5 = [ 12 × 0.5 × 2 ×[ 40/32.2] / 2= 7.47 lb. s/ft.
The viscous damping coefficient, c for 1.5= [ 12 × 1.5 × 2 ×[ 40/32.2] / 2 = 22.42 lb. s/ft.
Problem 10.012 SI A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at 2.6 bar, and saturated liquid exits the condenser at 12 bar. The isentropic compressor efficiency is 80%. The mass flow rate of refrigerant is 7 kg/min. Determine: (a) the compressor power, in kW. (b) the refrigeration capacity, in tons. (c) the coefficient of performance.
Answer:
a) 4.1 kw
b) 4.68 tons
c) 4.02
Explanation:
Saturated vapor enters compressor at ( p1 ) = 2.6 bar
Saturated liquid exits the condenser at ( p2 ) = 12 bar
Isentropic compressor efficiency = 80%
Mass flow rate = 7 kg/min
A) Determine compressor power in KW
compressor power = m ( h2 - h1 )
= 7 / 60 ( 283.71 - 248.545 )
= 4.1 kw
B) Determine refrigeration capacity in tons = m ( h1 - h4 )
= 7/60 ( 248.545 - 107.34 )
= 16.47 kw = 4.68 tons
C) coefficient of performance ( COP )
= Refrigeration capacity / compressor power
= 16.47 / 4.1 = 4.02
Attached below is the beginning part of the solution
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
A road is constructed at the capital cost of $6 million. At the end of Year 10, major improvements are to be made costing $17 million. At the end of Year 25, a replacement and upgrade is to be done at a cost of $29 million. At the end of year 40, the federal government issues a one-time tax credit in the amount of $12 million.
Over a 50-year analysis period (assuming a 10% interest rate) what is the annualized cost to the nearest dollar?
Answer:
The annualized cost is:
$299,272.
Explanation:
a) Data and Calculations:
Year 0 Capital cost of road construction = $6 million
Year 10 Major improvements cost = $17 million
Year 25 Replacement and upgrade cost = $29 million
Year 40 Federal government one-time tax credit = $12 million
Period of project analysis = 50 years
Cost of capital (discount rate) = 10%
Annualized cost at present value costs:
Amount spent Discount Factor Present value
$6 million 1 $6,000,000
$17 million 0.386 6,562,000
$29 million 0.092 2,668,000
($12 million) 0.0222 (266,400)
Total cost $14,963,600
Annualized cost = $14,963,600/50 = $299,272
b) The annualized cost for the road construction project, which is the annualized value of the net present costs of $14,963,600, is divided by 50. Before obtaining the net present costs, the cash outflows, including the tax credit, are discounted to their present values, using the discount rate of 10%. And then, the average of the cost is obtained by dividing the total net present cost into 50 years.
Technician A says that an A-pillar may be designed to transfer collision energy
Technician B says that a floor pan reinforcement may be designed to transfer collision energy
Who is right?
A. Aonly
B. B only
C. Both A and B
D. Neither Anor B
Technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.
What is Collision energy?Collision energy may be defined as a circumstance in which two or more bodies or particles come together with a resulting exchange of energy and alteration of direction.
Both pillar and floor pan reinforcement may be designed to transfer collision energy through the most prominent way to explain this mechanism of collisions.
A pillar exerts complete pressure on the floor in order to support the roof of the building, while floor pan reinforcement performs the same mechanism of collision energy with a different mechanism of action.
Therefore, technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.
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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.
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:
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
Which tool is used for cutting bricks and other masonry materials with precision?
Answer:
A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.
Explanation:
Answer:
Masonry Saw
Explanation:
Masonry Saws can be used to cut brick, ceramic, tile and or stone.
A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maximum shear stress is 450 N/cm2 when the maximum bending stress is 1500 N/cm2. Calculate the value of the centrally applied point load W and the span L. The overall height of the I section is 29 cm.
Answer:
W = 11,416.6879 N
L ≈ 64.417 cm
Explanation:
The maximum shear stress, [tex]\tau_{max}[/tex], is given by the following formula;
[tex]\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )[/tex]
[tex]t_w[/tex] = 1 cm = 0.01
h = 29 cm = 0.29 m
[tex]h_w[/tex] = 25 cm = 0.25 m
b = 15 cm = 0.15 m
[tex]I_c[/tex] = The centroidal moment of inertia
[tex]I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )[/tex]
[tex]I_c[/tex] = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴
Substituting the known values gives;
[tex]I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}[/tex]
[tex]I_c[/tex] = 1.2257083[tex]\bar 3[/tex] × 10⁻⁴ m⁴
From which we have;
[tex]4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )[/tex]
Which gives;
W = 11,416.6879 N
[tex]\sigma _{b.max} = \dfrac{M_c}{I_c}[/tex]
[tex]\sigma _{b.max}[/tex] = 1500 N/cm² = 15,000,000 N/m²
[tex]M_c[/tex] = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²
From Which we have;
[tex]M_{max} = \dfrac{W \cdot L}{4}[/tex]
[tex]L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417[/tex]
L ≈ 0.64417 m ≈ 64.417 cm.