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3 years ago

2) Why should you use a standard resistor so that the balance point is near the center of the bridge?

Engineering

1 answer:

Alina [70]3 years ago

7 0

Answer:

Explanation:

I think I will say it is because most, if not all resistors have been made to standard already.

The value for the resistor that is chosen for two fixed arms from the Wheatstone Bridge have to be quite bear each other, as much as possible. The arm that moves and the other that doesn't move, their resistance values are made to be quite close as well. This makes the resistance of the arm that moves to be set somewhere in the middle, and as a result, the measurements don't exceed beyond the resistance range of the arm that moves.

I hope you understand?

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Describe the energy transformation that occurs when a television is used

Morgarella [4.7K]

A television transforms electric energy into sound energy and radiant energy.

<h3>What is energy transformation?</h3>

Energy transformations are processes that convert energy from one type into another. Any type of energy use must involve some sort of energy transformation.

Some examples of energy transformations are, A television converts electric energy into sound and light energy. Lightbulbs convert electric energy into light energy. A hair dryer converts electrical energy into thermal and sound energy.

Learn more about Energy transformations here,

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2 years ago

The open-ended check is to check for ________.

Step2247 [10]

Answer:

any abnormalities

Explanation:

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4 years ago

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 maxi

saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, $\tau_{max}$ , is given by the following formula;

$\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )$

$t_w$ = 1 cm = 0.01

h = 29 cm = 0.29 m

$h_w$ = 25 cm = 0.25 m

b = 15 cm = 0.15 m

$I_c$ = The centroidal moment of inertia

$I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )$

$I_c$ = 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;

$I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}$

$I_c$ = 1.2257083 $\bar 3$ × 10⁻⁴ m⁴

From which we have;

$4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )$

Which gives;

W = 11,416.6879 N

$\sigma _{b.max} = \dfrac{M_c}{I_c}$

$\sigma _{b.max}$ = 1500 N/cm² = 15,000,000 N/m²

$M_c$ = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

$M_{max} = \dfrac{W \cdot L}{4}$

$L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417$

L ≈ 0.64417 m ≈ 64.417 cm.

4 0

3 years ago

How do scientists and engineers use math to help them?

VikaD [51]

Math (e.g., algebra, geometry, calculus, computer computation) is used both as a tool to create mathematical models that describe physical phenomena and as a tool to evaluate the merit of different possible solutions. ... In engineering, math and science are tools used within the engineering design process.
Biologists use math as they plot graphs to help them understand equations, run small “trial and error” tests with some sample numbers when developing algorithms, and use the R project for analyzing protein sequences and structures. Electrical engineers use math in many ways in their career. They use math to help design and test electrical equipment. They use math to calculate amp and volt requirements for electrical projects. They use math in creating computer simulations and designs for new products.

7 0

3 years ago

Water enters a hydraulic turbine through a 30-cm-diameter pipe at a rate of 0.6 m3s and exits through a 25-cm-diameter pipe. The

Vesnalui [34]

Answer:

Net electric power output = 54.594KW

Explanation:

Enter dia = 0.3m, Exit dia = 0.25, Flow Rate (Vflow) = 0.6m³/s, Hhg = 1.2m, Efficiency = 83%, Net electric power output = ?

Vflow = A1V1, where A1 is the Area of flow and V1 is the Velocity

V1 (Velocity at entering) = Vflow/A1 = 0.6/{(π/4)0.3²} = 8.48m/s

V2 (Velocity at exiting) = Vflow/A1 = 0.6/{(π/4)0.25²} = 12.22m/s

ΔP = (Shg - 1) x (ρh2o) x g x Hhg where Shg is the specific gravity of Mercury, ρh2o is the density of water, g is the acceleration due to gravity and Hhg is the height drop of the manometer

ΔP = (13.6 - 1) x 1000 x 9.81 x 1.2 = 148327.2Pa

Applying Bernoulli's Equation between the entering and exit

(P1/ρg) + α1({V1^2)/2g} + z1 = (P2/ρg) + α2({V2^2)/2g} + z2 + Hturbine

where z1, z2 = 0 as there no is change in the datum head and α is the correction factor = 1

Hturbine = (P1/ρg) - (P2/ρg) + α[{(Vq^2)/2g} - {(V2^2)/2g}]

Hturbine = (ΔP/ρg) + α[{(V1^2)/2g} - {(V2^2)/2g}]

Hturbine = (148327.2/1000 x 9.81) + 1[{(8.48^2)/2 x 9.81 - {(12.22^2)/2 x 9.81}] = 11.175m

The electrical Power Output is given by the equation Wturbine = η(turbine - generator) x ρ x Vflow x g x Hturbine

Wturbine = 0.83 x 1000 x 0.6 x 9.81 x 11.175 = 54.594 x 10^3 = 54.594 KW

3 0

4 years ago