Answer:
The Answer is C!
Explanation:
The pascal is actually a very small unit of pressure. To show this, convert 1 Pa = 1 N/m² to lb/ft². Atmosphere pressure at sea level is 14.7 lb/in². How many pascals is this?
Answer:
pascals is this = 101352.972 Pa
Explanation:
given data
Atmosphere pressure at sea level = 14.7 lb/in²
we convert 1 Pa = 1 N/m² to lb/ft²
so we convert here 14.7 lb/in² to pascals
we know that 1 lb/ft² = 47.990172 N/m²
so
1 lb/ft² × ft²/(12in)² = 47.990172 × 144 N/m²
it will be simplyfy
1 lb/ft² = 6894.76 N/m²
so
14.7 lb/in² = 14.7 × 6894.76 N/m²
14.7 lb/in² = 101352.972 Pa
Consider diodes in a rectifier circuit. Input voltage is sinusoidal with a peak of +/-10 V. Diode drop is 0.7 V. What is the PIV for each type rectifier 1. 0.7 V 2. 1.4 V 3. 10.7 V 4. 11.4 V Bridge rectifier 5. 19.3 V Full-wave rectifier 6. 8.6 V 7. 9.3 V Half-wave rectifier 8. 7.2 V 9. 12.1 V 10. 12.8 V 11. 10 V
Answer is given below:
Explanation:
Peak inverse voltage (PIV) can be defined as the maximum value of the reverse voltage of the diode, which is the maximum value of the input cycle when the diode is on. In reverse bias. Happens. 9.3V for braid rectifiers cut at 0.7The center tapered rectifier has 2 diodes in parallel so the maximum voltage is 2Vm so the answer to cut off the 0.7 voltage is19.3V. For a half wave rectifier it is Vm i.e. 10 V.Find the magnitude of the steady-state response of the system whose system model is given by dx(t)/dt+ x(t)-f(t), where f(t) 2cos8t. Keep 3 significant figures
This question is incomplete, the complete question is;
Find the magnitude of the steady-state response of the system whose system model is given by
dx(t)/dt + x(t) = f(t)
where f(t) = 2cos8t. Keep 3 significant figures
Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )
Explanation:
Given that;
dx(t)/dt + x(t) = f(t) where f(t) = 2cos8t
dx(t)/dt + x(t) = f(t)
we apply Laplace transformation on both sides
SX(s) + x(s) = f(s)
(S + 1)x(s) = f(s)
f(s) / x(s) = S + 1
x(s) / f(s) = 1 / (S + 1)
Therefore
transfer function = H(s) = x(s)/f(s) = 1/(S+1)
f(t) = 2cos8t → [ 1 / ( S + 1 ) ] → x(t) = Acos(8t - ∅ )
A = Magnitude of steady state output
S = jw
S = j8
so
A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )
A = 2/√65 = 0.2481
∅ = tan⁻¹( 1/1) = 45°
therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )
g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What is the friction factor
Answer:
The friction factor is 0.303.
Explanation:
The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:
[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)
Where:
[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.
[tex]D[/tex] - Diameter, measured in meters.
If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:
[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]
[tex]v \approx 5.093\,\frac{m}{s}[/tex]
The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:
[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)
If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:
[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]
[tex]Re = 211.230[/tex]
A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:
[tex]f = \frac{64}{Re}[/tex]
If we get that [tex]Re = 211.230[/tex], then the friction factor is:
[tex]f = \frac{64}{211.230}[/tex]
[tex]f = 0.303[/tex]
The friction factor is 0.303.
A uniform edge load of 500 lb/in. and 350 lb/in. is applied to the polystyrene specimen. If it is originally square and has dimensions of a = 2 in., b = 2 in., and a thickness of t = 0.25 in., determine its new dimensions a, b, and t after the load is applied. Ep = 597(10)3 psi, vp = 0.25.
The image of the load applied to the polystyrene is missing, so i have attached it.
Answer:
a_new = 2.00302 in
b_new = 2.00552
Explanation:
From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.
Now, formula for stress is;
Stress(σ) = Force/Area
We are not given force and area but the load and plate thickness.
Thus, stress = load/thickness
We are given;
Load in x - direction = 500 lb/in.
Load in y - direction = 350 lb/in.
Thickness; t = 0.25 in
Thus;
σ_x = 500/0.25
σ_x = 2000 ksi
σ_y = 350/0.25
σ_y = 1400 ksi
From Hooke's law for 2 dimensions, strain is given by the formula;
ε_x = (1/E)(σ_x - vσ_y)
ε_y = (1/E)(σ_y - vσ_x)
We are given v_p = 0.25 and Ep = 597 × 10³ psi
Thus;
ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)
ε_x = 0.00276
ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)
ε_y = 0.00151
From elongation formula, we know that;
Startin is: ε = ΔL/L
Thus; ΔL = Lε
We are given a = 2 and b = 2
Thus;
ΔL_x = 2 × 0.00276
ΔL_x = 0.00552
ΔL_y = 2 × 0.00151
ΔL_y = 0.00302
New dimensions are;
a_new = 2 + 0.00302
a_new = 2.00302 in
b_new = 2 + 0.00552
b_new = 2.00552
Calculate the normal stress along the x-direction.
[tex]\to \sigma_x =\frac{500}{t}=\frac{500}{0.25}= 2000 \ \frac{lb}{in^2}\\\\[/tex]
Calculate the normal stress along the y-direction.
[tex]\to \sigma_y =\frac{350}{t}=\frac{350}{0.25}= 1400 \ \frac{lb}{in^2}[/tex]
Calculate the strain along the x-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_y-\sigma_z)]\\\\[/tex]
[tex]=\frac{1}{597\times 10^3} [2000-0.25(1400+0)] \\\\= 2.764\times 10^{-3}\\\\[/tex]
Calculate the strain along the y-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_x+\sigma_z)][/tex]
[tex]=\frac{1}{597\times 10^3}[1400 -0.25 (2000+0)]\\\\=1508\times 10^{-3}[/tex]
Calculate the strain along the z-direction.
[tex]\to \varepsilon_x= \frac{1}{E}[\sigma_z-v(\sigma_x +\sigma_y)][/tex]
[tex]=\frac{1}{597\times 10^3} [0-0.25(2000+1400)] \\\\=-1.424\times 10^{-3}[/tex]
Calculate the new dimensions.
[tex]\to b' =b+ \varepsilon_xb\\\\[/tex]
[tex]= 2+2.764\times 10^{-3} \times 2\\\\= 2.005528 \ in\\\\[/tex]
[tex]\to a' = a + \varepsilon_y a\\\\[/tex]
[tex]= 2+1.508\times 10^{-3} \times 2\\\\= 2.003016\ in\\\\[/tex]
[tex]\to c'= c + \varepsilon_x c\\\\[/tex]
[tex]= 0.25 +(-1.424\times 10^{-3}) \times 0.25\\\\= 0.249644\ in\\\\[/tex]
Find out more about the dimensions here:
brainly.com/question/15406884
Compute the discharge observed at a v-notch weir. The weir has an angle of 90-degrees. The height above the weir is 3 inches.
Answer: the discharge observed at a v-notch weir is 66.7 in³/s
Explanation:
Given that;
Notch angle ∅ = 90°
height above the weir is 3 inches { head + head correction factor) h + k = 3 in
Discharge Q = ?
To determine the discharge observed, we us the following expression
Q = 4.28Ctan(∅/2) ( h + k )^5/2
where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor
now we substitute
Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2
Q = 4.28 × 1 × 1 × 15.5884
Q = 66.7 in³/s
Therefore the discharge observed at a v-notch weir is 66.7 in³/s
A layer of viscous fluid of constant thickness (no velocity perpenducilar to plate) flows steadily down an infinite, inclined plane. Determine, with the Navier Stokes equations, the flowrate per unit width as a function of flow height.
Answer:
q = (ρg/μ)(sin θ)(h³/3)
Explanation:
I've attached an image of a figure showing the coordinate system.
In this system: the velocity components v and w are equal to zero.
From continuity equation, we know that δu/δx = 0
Now,from the x-component of the navier stokes equation, we have;
-δp/δx + ρg(sin θ) + μ(δ²u/δy²) = 0 - - - - - (eq1)
Due to the fact that we have a free surface, it means we will not have a pressure gradient in the x-component and so δp/δx = 0
Then our eq 1 is now;
ρg(sin θ) + μ(δ²u/δy²) = 0
μ(δ²u/δy²) = -ρg(sin θ)
Divide both sides by μ to get;
(δ²u/δy²) = -(ρg/μ)(sin θ)
Integrating both sides gives;
δu/δy = -(ρg/μ)(sin θ)y + b1 - - - - (eq2)
Now, the shear stress is given by the formula;
τ_yx = μ[δu/δy + δv/δx]
From the diagram, at the free surface,τ_yx = 0 and y = h
This means that δu/δy = 0
Thus, putting 0 for δu/δy in eq 2, we have;
0 = -(ρg/μ)(sin θ)h + b1
b1 = h(ρg/μ)(sin θ)
So, eq 2 is now;
δu/δy = -(ρg/μ)(sin θ)y + h(ρg/μ)(sin θ)
Integrating both sides gives;
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y + b2 - - - eq3
Because u = 0 when y = 0, it means that b2 = 0 also because when we plug 0 for u and y into eq3, we will get b2 = 0.
Thus, we now have:
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y
Factorizing like terms, we have;
u = (ρg/μ)(sin θ)[hy - y²/2] - - - (eq 4)
The flow rate per unit width is gotten by Integrating eq 4 between the boundaries of h and 0 to give;
∫u = (h,0)∫(ρg/μ)(sin θ)[hy - y²/2]
q = (ρg/μ)(sin θ)[hy²/2 - y³/6] between h and 0
q = (ρg/μ)(sin θ)[h³/2 - h³/6]
q = (ρg/μ)(sin θ)(h³/3)
While at a concert you notice five people in the crowd headed in the same direction. Your tendency to group them is due to? *
Answer:
common fate
Explanation:
The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.
Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.
Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.