Answer:
Technician B
Explanation:
In simplistic terms, the "hot wire" connects the load device to the source of electrical energy. The ground wire provides the return path for current from the load device to the energy source. In many circuits, the "ground wire" is at, near, or defined as "ground" potential (the actual potential of the Earth).
Technician A seems to be confused. Technician B is more correct.
Which of the following is NOT a factor that contributes to the annual cost to own an automobile
a
vehicle color
b
repairs
c
fuel
d
maintenance
Answer:
a. vehicle color
Explanation:
One might expect that the cost of owning a vehicle would not be a function of its color. There is no reason to believe that a blue car gets better gas mileage than a green car of the same make, model, and equipment. So, the appropriate choice is ...
a. vehicle color
__
However, vehicle color may play a role in ownership costs if more money is spent on washing a white car than would be spent on a black or beige car. Similarly, a light-colored car may require less use of an air-conditioner in the summer sun than does a dark-colored car, ultimately affecting fuel cost. It isn't always obvious what the features of a vehicle are that contribute to ownership cost.
what was the first roblox player A.bob B.roblox C.builderman
Answer:
actually, it was John Doe and Jane Doe, as they were created as testers on the exact same say when Roblox was created.
Explanation:
The first player was builderman. Option C
Who was the first player?The first player was one whose user handle was builderman.
However, the account was terminated. It was replaced with
‘Builderman’ is the account that belonged to the CEO and Co-Founder of , David Baszucki. He manages all of the admins.
When you make a new account, he's automatically your friend.
Note that ‘John Doe’ and ‘Jane Doe’ were two test accounts that were created by David.
The accounts were later rumored to hack others’
Learn about username at: https://brainly.com/question/28344005
#SPJ6
Consider the function represented by the equation x – y = 3. What is the equation written in function notation, with x as the independent variable?
Answer:
f(x) = x -3
Explanation:
Solve for y, then replace y by f(x).
x - y = 3
x - 3 = y . . . . . add y-3 to both sides
f(x) = x -3
Answer:
(x) = x – 3
Explanation:
edge
Shania has started a new job as an app developer. Her first task was to make an old designed for Android available on other platforms. which of the following would make her job easiest and fastest?
A storage tank has a volume of 3,000 liters. The tank is originally filled to a pressure of 21.1 megapascals with an ideal gas while the temperature is maintained at 28 degrees Celsius. The tank is heated so the temperature increases to a final temperature of 58 degrees Celsius. What will be the new pressure in the tank in units of megapascals?
It is given that volume is constant i.e 3000 L.
So, [tex]\dfrac{T_1}{P_1}=\dfrac{T_2}{P_2}[/tex]
Now,
[tex]T_1=28 +273 = 301\ K\\\\T_2=58 + 273 = 331\ K\\\\P_1 = 21.1\ MPa[/tex]
Putting all values in above equation, we get :
[tex]\dfrac{301}{21.1}=\dfrac{331}{P_2}\\\\P_2=\dfrac{331\times 21.1}{301}\ MPa\\\\P_2=23.20\ MPa[/tex]
Therefore, the new pressure will be 23.20 MPa.
Hence, this is the required solution.
. On Juan's twenty-sixth birthday, he deposited $7,500 in a retirement account. Each year thereafter, he deposited $1,000 more than the previous year. Using a gradient series factor, determine how much was in the account immediately after his thirty-fifth deposit if
Answer:
$1,783,805
Explanation:
Okay, if we are going to answer thus question efficiently, we need to take time to understand each and every sentence in the question. Hence, from the question we have the following data or information and they are;
1. The initial deposit in which Juan deposited on his 26th birthday = $7500.
2. After the statement in [1] above, Juan decided to be depositing $1000 more than the one he deposited on his 26th birthday.
Therefore, we are given that there is 5% in the interest rate. Thus, after his thirty-fifth deposit he will have;
$7500[90.3203] + [5.516] × $1000 × 200.581] = $1,783,805
2
A spring balance pulls with 5 N on a beam of 0.5 m.
What is the torque at the end of the beam?
Answer:
The torque at the end of the beam is 2.5 Nm
Explanation:
Given;
length of beam, r = 0.5 m
applied force, F = 5 N
The torque at the end of the beam is given by;
τ = F x r
where;
τ is the torque
F is applied force
r is length of the beam
τ = 5 x 0.5
τ = 2.5 Nm
Therefore, the torque at the end of the beam is 2.5 Nm
Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is −5°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 = 10 W/m2⋅K and h2 = 25 W/m2⋅K, respectively, and disregard any heat transfer by radiation.
Answer:
The steady rate of heat transfer through the glass window is 707.317 watts.
Explanation:
A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:
[tex]\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}[/tex] (Eq. 1)
Given that window is represented as a flat element, we can expand (Eq. 1) as follows:
[tex]\dot Q_{total} = \frac{T_{i}-T_{o}}{R}[/tex] (Eq. 2)
Where:
[tex]T_{i}[/tex], [tex]T_{o}[/tex] - Indoor and outdoor temperatures, measured in Celsius.
[tex]R[/tex] - Overall thermal resistance, measured in Celsius per watt.
Now, we know that glass window is configurated in series and overall thermal resistance is:
[tex]R = R_{cond} + R_{conv, in}+R_{conv, out}[/tex] (Eq. 3)
Where:
[tex]R_{cond}[/tex] - Conductive thermal resistance, measured in Celsius per watt.
[tex]R_{conv, in}[/tex], [tex]R_{conv, out}[/tex] - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.
And we expand the expression as follows:
[tex]R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}[/tex]
[tex]R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)[/tex] (Eq. 4)
Where:
[tex]w[/tex] - Width of the glass window, measured in meters.
[tex]d[/tex] - Length of the glass window, measured in meters.
[tex]l[/tex] - Thickness of the glass window, measured in meters.
[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius.
[tex]h_{i}[/tex], [tex]h_{o}[/tex] - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.
If we know that [tex]w = 2.4\,m[/tex], [tex]d = 1.5\,m[/tex], [tex]l = 0.006\,m[/tex], [tex]k = 0.78\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex] and [tex]h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex], the overall thermal resistance is:
[tex]R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)[/tex]
[tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex]
Now, we obtain the steady rate of heat transfer from (Eq. 2): ([tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex], [tex]T_{i} = -5\,^{\circ}C[/tex], [tex]T_{o} = 24\,^{\circ}C[/tex])
[tex]\dot Q_{total} = \frac{24\,^{\circ}C-(-5\,^{\circ}C)}{0.041\,\frac{^{\circ}C}{W} }[/tex]
[tex]\dot Q_{total} = 707.317\,W[/tex]
The steady rate of heat transfer through the glass window is 707.317 watts.
A pump is positioned at 2 m above the water of the reservoir. The inlet of the pipe connected to the pump is positioned at 6m beneath the water of the reservoir. When a pump draws 220 m3/hour of water at 20 °C from a reservoir, the total friction head loss is 5 m. The diameter of the pipe connected to the inlet and exit nozzle of the pump is 12 cm and 5 cm, respectively. The flow discharges through the exit nozzle to the atmosphere. Calculate the pump power in kW delivered to the water.
Answer:
The pump delivers 32.737 kilowatts to the water.
Explanation:
We can describe the system by applying the Principle of Energy Conservation and the Work-Energy Theorem, the pump system, which works at steady state and changes due to temperature are neglected, is represented by the following model:
[tex]\dot W_{in} + \dot m \cdot g \cdot (z_{1}-z_{2}) + \frac{1}{2}\cdot \dot m \cdot (v_{1}^{2}-v_{2}^{2})+\dot m \cdot [(u_{1}+P_{1}\cdot \nu_{1})-(u_{2}+P_{2}\cdot \nu_{2})]-\dot E_{losses} = 0[/tex] (Eq. 2)
Where:
[tex]\dot m[/tex] - Mass flow, measured in kilograms per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final flow speeds at pump nozzles, measured in meters per second.
[tex]u_{1}[/tex], [tex]u_{2}[/tex] - Initial and final internal energies, measured in joules per kilogram.
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.
[tex]\nu_{1}[/tex], [tex]\nu_{2}[/tex] - Initial and final specific volumes, measured in cubic meters per kilogram.
Then, we get this expression:
[tex]\dot W_{in} + \dot m \cdot g \cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot \dot m \cdot (v_{1}^{2}-v_{2}^{2}) +\dot m\cdot \nu \cdot (P_{1}-P_{2})-\dot E_{losses} = 0[/tex] (Ec. 3)
We note that specific volume is the reciprocal of density:
[tex]\nu = \frac{1}{\rho}[/tex] (Ec. 4)
Where [tex]\rho[/tex] is the density of water, measured in kilograms per cubic meter.
The initial pressure of water ([tex]P_{1}[/tex]), measured in pascals, can be found by Hydrostatics:
[tex]P_{1} = P_{atm} + \rho\cdot g \cdot \Delta z[/tex] (Ec. 5)
Where:
[tex]P_{atm}[/tex] - Atmospheric pressure, measured in pascals.
[tex]\Delta z[/tex] - Depth of the entrance of the inlet pipe with respect to the limit of the water reservoir.
If we know that [tex]p_{atm} = 101325\,Pa[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta z = 6\,m[/tex], then:
[tex]P_{1} = 101325\,Pa+\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}})\cdot (6\,m)[/tex]
[tex]P_{1} = 160167\,Pa[/tex]
And the specific volume of water ([tex]\nu[/tex]), measured in cubic meters per kilogram, is: ([tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex])
[tex]\nu = \frac{1}{1000\,\frac{kg}{m^{3}} }[/tex]
[tex]\nu = 1\times 10^{-3}\,\frac{m^{3}}{kg}[/tex]
The power losses due to friction is found by this expression:
[tex]\dot E_{losses} = \dot m \cdot g\cdot h_{losses}[/tex]
Where [tex]h_{losses}[/tex] is the total friction head loss, measured in meters.
The mass flow is obtained by this:
[tex]\dot m = \rho \cdot \dot V[/tex] (Ec. 6)
Where [tex]\dot V[/tex] is the volumetric flow, measured in cubic meters per second.
If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex] and [tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], then:
[tex]\dot m = \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(0.061\,\frac{m^{3}}{s} \right)[/tex]
[tex]\dot m = 61\,\frac{kg}{s}[/tex]
Then, the power loss due to friction is: ([tex]h_{losses} = 5\,m[/tex])
[tex]\dot E_{losses} = \left(61\,\frac{kg}{s}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (5\,m)[/tex]
[tex]\dot E_{losses} = 2991.135\,W[/tex]
Now, we calculate the inlet and outlet speed by this formula:
[tex]v = \frac{\dot V}{\frac{\pi}{4}\cdot D^{2} }[/tex] (Ec. 7)
Inlet nozzle ([tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], [tex]D = 0.12\,m[/tex])
[tex]v_{1} = \frac{0.061\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.12\,m)^{2} }[/tex]
[tex]v_{1} \approx 5.394\,\frac{m}{s}[/tex]
Oulet nozzle ([tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], [tex]D = 0.05\,m[/tex])
[tex]v_{2} = \frac{0.061\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.05\,m)^{2} }[/tex]
[tex]v_{2} \approx 31.067\,\frac{m}{s}[/tex]
([tex]\dot m = 61\,\frac{kg}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2} = 2\,m[/tex], [tex]z_{1} = -6\,m[/tex], [tex]v_{2} \approx 31.067\,\frac{m}{s}[/tex], [tex]v_{1} \approx 5.394\,\frac{m}{s}[/tex], [tex]P_{2} = 101325\,Pa[/tex], [tex]P_{1} = 160167\,Pa[/tex], [tex]\dot E_{losses} = 2991.135\,W[/tex])
[tex]\dot W_{in} = \left(61\,\frac{kg}{s}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [2\,m-(-6\,m)]+\frac{1}{2}\cdot \left(61\,\frac{kg}{s}\right) \cdot \left[\left(31.067\,\frac{m}{s} \right)^{2}-\left(5.394\,\frac{m}{s} \right)^{2}\right] +\left(61\,\frac{kg}{s}\right)\cdot \left(1\times 10^{-3}\,\frac{m^{3}}{kg} \right)\cdot (101325\,Pa-160167\,Pa)+2991.135\,W[/tex]
[tex]\dot W_{in} = 32737.518\,W[/tex]
The pump delivers 32.737 kilowatts to the water.
What are the four different types of auto body shops and three different positions within the shop?
Types; 2 Auto body repair; 3 Auto glass repair; 4 See also; 5 References ... Independent automobile repair shops in the US may also achieve certification through manufacturer sponsored ...
A stone is catapulted at time t = 0, with an initial velocity of magnitude 21.1 m/s and at an angle of 39.9° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.08 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.75 s, and for the (e) horizontal and (f) vertical components at t = 5.09 s. Assume that the catapult is positioned on a plain horizontal ground.
Answer:
(a) X = 17.48 m
(b) Y = 8.903 m
(c) X = 28.33 m
(d) Y = 8.68 m
(e) X = 44.68 m
(f) Y = 0 m
Explanation:
Given;
magnitude of initial velocity, v = 21.1 m/s
angle of projection, θ = 39.9°
the horizontal component of the velocity, [tex]v_x = vcos \theta[/tex]
[tex]v_x = 21.1(cos 39.9^0) = 16.187 \ m/s[/tex]
the vertical component of the velocity, [tex]v_y = vsin \theta[/tex]
[tex]v_y = 21.1(sin 39.9^0) = 13.535 \ m/s[/tex]
(a) the horizontal components of its displacement;
[tex]X = v_xt + \frac{1}{2} gt^2[/tex]
gravitational influence on horizontal direction is negligible, g = 0
[tex]X = v_xt \\\\X = (16.187)(1.08)\\\\X = 17.48 \ m[/tex]
(b) the vertical components of its displacement;
[tex]Y = v_yt + \frac{1}{2} gt^2\\\\Y = (13.535*1.08) + \frac{1}{2} (-9.8)(1.08)^2\\\\Y = 14.618 -5.715\\\\Y = 8.903 \ m\\[/tex]
(c) the horizontal components of its displacement;
[tex]X = v_xt\\\\ X = (16.187)(1.75)\\\\X = 28.33 \ m[/tex]
(d) the vertical components of its displacement;
[tex]Y = v_yt + \frac{1}{2} gt^2\\\\Y = (13.535 *1.75) + \frac{1}{2} (-9.8)(1.75)^2\\\\Y = 8.68 \ m[/tex]
(e) the horizontal components of its displacement;
[tex]X = v_xt \\\\X = (16.187*5.09)\\\\X = 82.39 \ m\\[/tex]
(f) the vertical components of its displacement;
[tex]Y = v_yt + \frac{1}{2}gt^2\\\\ Y = (13.535*5.09)+ \frac{1}{2}(-9.8)(5.09)^2\\\\Y = -58.1 \ m[/tex]
this displacement is not possible because it is beyond zero vertical level; it shows that the stone will hit the ground before 5.09 s.
When the stone hits the ground Y = 0 m
At zero vertical displacement (Y = 0 m), the time at that position will be calculated as;
[tex]Y = v_yt+\frac{1}{2}gt^2\\\\0 = (13.535t) + \frac{1}{2}(-9.8)t^2\\\\0= 13.535t - 4.9t^2\\\\0 = t(13.535-4.9t)\\\\t = 0 \ \ or \\\\13.535-4.9t = 0\\\\4.9t = 13.535\\\\t = 2.76 \ s[/tex]
The horizontal displacement at this time is given by;
X = vₓt
X = (16.187 x 2.76)
X = 44.68 m
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate which is initially at a uniform temperature of Ti = 200 degree C and is to be heated to a minimum temperature of 550 degree C. Heating is effected in a gas-fired furnace, where products of combustion at T infinity = 800 degree C maintain a convection coefficient of h = 250 W/m2 K on both surfaces of the plate How long should the plate be left in the furnace?
Complete question is;
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?
Answer:
860 seconds
Explanation:
We are given;
Initial Temperature; Ti = 200 °C
Minimum Temperature; T_i = 550 °C
T∞ = 800 °C
convection coefficient; h = 250 W/m².K
ρ = 7830 kg/m³
Cp = 550 J/kg K
k = 48 W/m K
Plate thickness = 100mm
Thus,L = 100/2 = 50mm = 0.05 m
Let's find the biot number from the formula;
Bi = hL/K
Bi = (250 × 0.05)/48
Bi = 0.2604
Now, lowest temperature in the slab is given as;
θ_o = (T_o - T∞)/(T_i - T∞)
θ_o = (550 - 800)/(200 - 800)
θ_o = 0.4167
Now, from online tables calculation, we can find the root of the biot number.
Thus, root of the biot number Bi = 0.2604 is;
ζ1 = 0.488 rad
Also, C1 is gotten as 1.0396
Now,formula for thermal diffusivity is;
α = k/ρc
α = 48/(7830 × 550)
α = 1.115 × 10^(-5) m²/s
Also, from online tables, f(ζ1) = 0.401
Thus, we can find the time the plate should the plate be left in the furnace from;
-(ζ1)²(αt/L²) = In 0.401
-(ζ1)²(αt/L²) = -0.9138
t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))
t ≈ 860 s
In this exercise we want to calculate the time, in seconds, of the time left on the plate in the furnace, like this:
860 seconds
Organizing the information given in the statement we find that:
Initial Temperature; Ti = 200 °CMinimum Temperature; T_i = 550 °CInfinity Temperature: T=800°Cconvection coefficient; h = 250 W/m².Kρ = 7830 kg/m³Cp = 550 J/kg Kk = 48 W/m KPlate thickness = L = 0.05 mUsing the formula given below, we will have how to calculate the number of biot, like this:
[tex]B = hL/K\\B = (250 * 0.05)/48\\B = 0.2604[/tex]
calculating the angle that corresponds to the temperature difference as:
[tex]\theta_o = (T_o - T)/(T_i - T)\\\theta_o = (550 - 800)/(200 - 800)\\\theta_o = 0.4167[/tex]
Using the formula below, we will have:
[tex]\alpha = k/\rho c\\\alpha = 48/(7830 * 550)\\\alpha = 1.115 * 10^{-5}[/tex]
Combining all the information from the previous calculations, we have that the time will be calculated as:
[tex]-(\phi)^2(\alpha t/L^2) = In 0.401\\-(\phi )^2(\apha t/^2) = -0.9138\\t = (-0.9138 * 0.05^2)/-(0.488^2 * 1.115 * 10^{-5})\\t = 860 s[/tex]
See more about time at brainly.com/question/2570752
In Female, the twenty-third pair of chromosomes is called in in
A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is 0.0500. Liquid hexane condensate is recovered at a rate of 1.50 L/min. (a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.) (b) What percentage of the hexane entering the condenser is recovered as a liquid?
Answer:
A. 72.53 mol/min
B. 76.2%
Explanation:
Please check attachment to see the diagram of the components I drew.
I first converted flow rate to a molar flow rate. I got 11.5mol/ min after converting. Check attachment
A.) When we create a balance on all these components
n1 + n2 + n3
n1 = n2 + 11.5
Creating a balance on hexane
n1y1C6H14 = n2y2C6H14 + n3y3C6H14
n1(0.18) = n2(0.05) + 11.5mol/min(1.00)
When we put the first equation into the second one:
(n2+11.5mol/min)0.18 + n2(0.05) + 11.5mol/min(1.00)
= 0.18n2 + 2.07min/mol = 0.5n2 + 11.5mol/min
Collect like terms
0.18n2 - 0.05n2 = 11.5 - 2.07min/mol
= 0.13n2 = 9.43
n2 = 9.43/0.13
n2 = 72.53 mol/min
72.53 is the flow rate of the gas stream leaving the burner.
B.) n1 = n2 + 11.5mol/min
n1 = 72.53mol/min + 11.5mol/min
n1 = 84.93mol/min
Amount of hexane entering condenser
n1c6H14 = y1C6H14 = 0.18(84.03mol/min)
= 15.1mol/min
The percentage condensed
= (11.5mol/min)/(15.1mol/min) * 100
= 76.2% of hexane is recovered as liquid
How is the foundation for a skyscraper different from a house?
Answer:
Shallow foundations, often called footings, are usually embedded about a metre or so into soil. ... Another common type of shallow foundation is the slab-on-grade foundation where the weight of the structure is transferred to the soil through a concrete slab placed at the surface.
Explanation:
Because I said so.
3.94 x 105) + (2.04 x 105)
what is tcp in network?
Hammer welding preceded resistance welding
True
False
Answer:false
Explanation:
Bc
Which of the following best describes empathy?
the understanding of the feelings and beliefs of others
the lack of pride or boastfulness
the courage to speak up with one’s ideas
the possession of honesty and high morals
Answer:
the first one is the correct answer
Answer:
the first one would be correct
Explanation:
Heeeelppp please um if your smart help
Answer:
ben black diyorum ve Iyi srdsler dilerim
#elmaliturta23
Find the volume of the rectangular prism
9 cm
10 cm
Answer:
V= 90h cm³ where h is the height of the rectangular prism.
Explanation:
The formula for volume of a rectangular prism is ;
V=l*w*h where;
V=volume in cm³
l= length of prism=10cm
w =width of the prism = 9 cm
Assume the height of the prism as h cm then the volume will be;
V= 10* 9*h
V= 90h cm³
when the value of height of the prism is given, substitute that value with h to get the actual volume of the prism.
Think about a good game story that made you feel a mix of positive and negative emotions. What was the story, what emotions did you feel, and how did it make you feel them? Why did those emotions draw you into the story?
What is a computer ? Does it work or not I’m asking this Bc I’m stupid
Answer:
A computo is a box and it works really well if its good
the thurst from a bottle rocket last until
Answer:
what are you in third grade??
Explanation:
the answer is the air pressure inside of the bottle reaches the same air pressure as outside of the bottle. K?? got it love???I hope soo
Answer:
the air pressure inside of the bottle reaches the same air pressure as outside of the bottle.
Explanation:
No help dude that’s not even part of the question
Answer:
wat?
Explanation:
The question should be labeled in the psychology section, but what I assume the question means is some sort of paradoxical reverse psychology method of braincell loss. Even photosynthesis doesn't understand what it means. The question probably is what is the question, not that there's no help.
Hope this helps (a little)!