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svet-max [94.6K]
3 years ago
10

Liquidated damages are intended to represent anticipated losses to the owner based upon circumstances existing at the time the c

ontract was made. List at least five types of potential losses to the owner that would qualify for determination of such potential losses.
Engineering
1 answer:
Ganezh [65]3 years ago
4 0

Answer:

1. Loss of income.

2. Rental costs.

3. Utility bills.

4. Loss of rent.

5. Storage costs.

Explanation:

Liquidated damages can be defined as pre-determined damages or clauses that are highlighted or indicated at the time of entering into a contract between a contractor and a client which is mainly based on evaluation of the actual loss the client may incur should the contractor fail to meet the agreed completion date.

Generally, liquidated damages are meant to be fair rather than being a penalty or punitive to the defaulter. It is usually calculated on a daily basis for the loss.

When entering into a contract with another, liquidated damages are intended to represent anticipated losses to the owner based upon circumstances existing at the time the contract was made.

Listed below are five (5) types of potential losses to the owner that would qualify for determination of such potential losses;

1. Loss of income.

2. Rental costs.

3. Utility bills.

4. Loss of rent.

5. Storage costs.

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If you know that the change in entropy of a cup of coffee where heat was added is 20 J/K, and that the temperature of the coffee
Anna35 [415]

Answer:

5000J

Explanation:

Given in the question that

Heat added to the coffee cup is, ΔS = 20 J/K

The temperature of the coffee, T = 250 K

Now, using the formula for the entropy change

\bigtriangleup S=-\frac{\bigtriangleup H}{T} ...........(1)

Where,

ΔS is the entropy change

ΔH is the enthalpy change

T is the temperature of the system

substituting the values in the equation (1)

we get

20=-\frac{\bigtriangleup H}{250}

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ΔH=5000 J

6 0
3 years ago
The hypotenuse of a 45° right triangle is
wlad13 [49]
75+69 and divide by 54
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3 years ago
What is the greatest speedup (in terms of throughput) possible with pipelining? You do not need to worry about the register timi
RUDIKE [14]

Answer:

The greatest speedup possible with pipelining is 2.5 times

Explanation:

As we know

Span of stage = Maximum span of any span

As stage B has the maximum duration of time, So

Duration of time of each stage = Time duration of Stage B = 8 ns

Now calculate the total time without pipeline = Duration of Stage A + Duration of Stage B + Duration of Stage C + Duration of Stage D = 5 ns + 8 ns + 4 ns + 3 ns = 20 ns

So the greatest speedup can be calculated as follow

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3 0
3 years ago
Two steel plates of 15 mm thickness each are clamped together with an M14 x 2 hexagonal head bolt, a nut, and one 14R metric pla
Blababa [14]

Answer:

a) 50 mm

b) 808.24 MN/m

Explanation:

Given:

Thickness of each steel plate = 15mm

a) To find suitable length of bolt, we'll use:

Length of bolt = grip length + height of nut

To find the grip length since there is a washer, we'll use:

Grip length = plate thickness + washer thickness

Since we have 2 plates of 15mm thickness,

Plate thickness = 15 + 15 = 30mm

Using the table, a metric plain washer has a thickness of 3.5mm

Grip length = 30 + 3.5 =33.5 mm

Height of nut: Using table A-31, height of hexagonal nut is 12.8 mm

Therefore,

Length of bolt = 33.5 + 12.8 = 46.3

Rounde up to the nearest 5mm, we'll get 50mm

Length of bolt = 50mm

b) Bolt stiffness:

Threaded length for L ≤ 125mm

LT = 2*d + 6

Where d = 14mm, from table(8-7)

= 2*14 + 6

= 34 mm

Area of unthreaded portion:

Ad= \frac{\pi}{4} d^2 = \frac{pi}{4} * 14^2 = 153.94 mm^2

Length of unthreaded portion in grip:

Ld = 50 - 34 = 16mm

Length of threaded portion in grip:

Lt = 33.5 - 16 = 17.5mm

Bolt stiffness = \frac{A_d A_t E}{A_d l_t + A_t l_d}

= \frac{153.94 * 115 * 207}{(153.94*17.5)+(115*16)} = 808.24

Bolt stiffness = 808.24 MN/m

7 0
3 years ago
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