The best way to identify common masonry problems is to call the engineer.<br> True or False

A photovoltaic panel of dimension 2mÃ—4m is installed on the

blsea [12.9K]

Answer:

Explanation:

Simply put, a solar panel works by allowing photons, or particles of light, to knock electrons free from atoms, generating a flow of electricity. Solar panels actually comprise many, smaller units called photovoltaic cells. (Photovoltaic simply means they convert sunlight into electricity. The attached diagram give an ilustsration of the photovotaic pannel mounted on a roof top.

Solution

To Determine the electric power generated for

a) A still summer day.

E = A * r * H * PR

E = Total Amount of Energy in kilowatt

A = Total Surface Area

r = efficiency Rating

H = global radiation value

PR = Performance Ratio

kwh = watt * Time/1000

kwh = 100 * 35/1000

3.5

b)

kwh = watt * Time/1000

kwh = 30 *15/1000

4.5

6 0

3 years ago

Pendulum impacting an inclined surface of a block attached to a spring-Dependent multi-part problem assign all parts NOTE: This

Art [367]

Answer:

vA = -2.55 m/s

vB = 0.947 m/s

Explanation:

Given:-

- The initial angle of rope, α = 30°

- The angle of rope just before impact or wedge angle, θ = 20°

- The weight of sphere, Ws = 1-lb

- The initial position velocity, vi = 4 ft/s

- The coefficient of restitution, e = 0.7

- The weight of the wedge, Ww = 2-lb

- The spring constant, k = 1.8 lb/in

- The length of rope, L = 2.6 ft

Find:-

Determine the velocities of A and B immediately after the impact.

Solution:-

- We can first consider the ball ( acting as a pendulum ) to be isolated for study.

- There are no unbalanced fictitious forces acting on the sphere ball. Hence, we can reasonably assume that the energy is conserved.

- According to the principle of conservation for the initial point and the point just before impact.

Let,

vA : The speed of sphere ball before impact

Change in kinetic energy = Change in potential energy

ΔK.E = ΔE.P

0.5*ms* ( uA^2 - vi^2 ) = ms*g*L*( cos ( θ ) - cos ( α ) )

uA^2 = 2*g*L*( cos ( θ ) - cos ( α ) ) + vi^2

uA = √ [ 2*32*2.6*( cos ( 20 ) - cos ( 30 ) ) + 4^2 ] = √28.25822

uA = 5.316 ft/s

- The coefficient of restitution (e) can be thought of as a measure of the extent to which mechanical energy is conserved when an object bounces off a surface:

e^2 = ( K.E_after impact / K.E_before impact )

- The respective Kinetic energies are:

K.E_after impact = K.E_sphere + K.E_block

= 0.5*ms*vA^2 + 0.5*mb*vB^2

K.E_before impact = K.E = Ws*L*( cos ( θ ) - cos ( α ) )

= 1*2.6*( cos ( 20 ) - cos ( 30 ) )

= 0.1915 J

32*2*0.1915*0.7^2 = Ws*vA^2 + Wb*vB^2

6.00544 = vA^2 + 2*vB^2 ... Eq1

- From conservation of linear momentum we have:

vB = e*( uA - uB )*cos ( 20 ) + vA

vB = 0.7*( 5.316 - 0 )*cos ( 20) + vA

vB = 3.49678 + vA .... Eq 2

- Solve two equation simultaneously:

6.00544 = vA^2 + 2*(3.49678 + vA)^2

6.00544 = 3vA^2 + 13.98*vA + 24.455

3vA^2 + 14.8848*vA + 18.4495 = 0

vA = -2.55 m/s

vB = 0.947 m/s

5 0

4 years ago

You could be sued if you injure someone while rescuing them if...

navik [9.2K]

I believe number 4 I could be wrong but I think it’s 4

3 0

3 years ago

Read 2 more answers

Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relat

Serggg [28]

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

Eq 3

Finally, solving for unknown pressure as a function of hoop strain:

Eq 4

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

Eq 8

For Wheat stone bridge:

Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

Eq 11

Download docx

3 0

3 years ago

Risks are Not Perceived Differently from What is Happening<br> False<br> True

Gnoma [55]

Answer:

False......................

8 0

3 years ago

The best way to identify common masonry problems is to call the engineer. True or False