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3 years ago

Why is the contractor normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract?

Engineering

1 answer:

just olya [345]3 years ago

3 0

Answer:

It serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

Explanation:

A bid bond is a type of construction bond that protects the obligee in a construction bidding process.

A bid bond typically involves three parties:

The obligee; the owner or developer of the construction project under bid. The principal; the bidder or proposed contractor.

The surety; the agency that issues the bid bond to the principal example insurance company or bank.

A bid bond generally serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

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A Scalar can only be a positive quantity that has a magnitude but no direction ? a)-True b)-False

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4 years ago

Air at a temperature of 333 °C flows with a velocity of 10 m/s over a flat plate 0.5m long. Estimate the cooling rate per unit w

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4 years ago

Jeff is assembling a rear axle unit. He has all

BartSMP [9]

Answer:

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3 years ago

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3 years ago

Water flows through a multisection pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s a

Alex_Xolod [135]

Answer:

b. 2.3 kPa.

Explanation:

This situation can be modelled by Bernoulli's Principle, as there are no energy interaction throughout the multisection pipe and current lines exists between both ends. Likewise, this system have no significant change in gravitational potential energy since it is placed horizontally on the ground and is described by the following model:

$P_{1} + \rho \cdot \frac{v_{1}^{2}}{2} = P_{2} + \rho \cdot \frac{v_{2}^{2}}{2}$

Where:

$P_{1}$ , $P_{2}$ - Pressures at the beginning and at the end of the current line, measured in kilopascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Fluid velocity at the beginning and at the end of the current line, measured in meters per second.

Now, the pressure difference between these two points is:

$P_{1} - P_{2} = \rho \cdot \frac{v_{2}^{2}-v_{1}^{2}}{2}$

If $\rho = 1000\,\frac{kg}{m^{3}}$ , $v_{1} = 3\,\frac{m}{s}$ and $v_{2} = 2.1\,\frac{m}{s}$ , then:

$P_{1} - P_{2} = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \frac{\left(2.1\,\frac{m}{s} \right)^{2}-\left(3\,\frac{m}{s} \right)^{2}}{2}$

$P_{1} - P_{2} = -2295\,Pa$

$P_{1} - P_{2} = -2.295\,kPa$ (1 kPa is equivalent to 1000 Pa)

Hence, the right answer is B.

7 0

3 years ago