Explain the 11 sections that a typical bill of quantity is divided into

A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int

Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

$s = v_0t + \frac{at^2}{2}$

$s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)$

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0

2 years ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

6 0

2 years ago

What substance does light travel through before putting water in the cup

torisob [31]

Bend surface in water! Hopefully this helps, I looked it up!

6 0

2 years ago

A source outputs data at the rate of 1,000,000 bits/sec. The transmitter uses binary PAM with raised cosine pulse shaping. If th

musickatia [10]

Answer:

maximum roll-off factor is 0.7

Explanation:

given data

rate of output data = 1,000,000 bits/sec

cutoff frequency fc = 850 kHz

solution

we know bandwidth frequency is same as that of bandwidth

we apply here fc formula that is

fc = ω ( 1 + α ) ..........1

here

ω = data rate ÷ 2 .............2

ω = 1000 ÷ 2

ω = 500000 Hz

so put now value in eq 1

fc = ω ( 1 + α )

850000= 500000 ( 1 + α )

solve it we get

α = 0.7

so, maximum roll-off factor is 0.7

7 0

2 years ago

250 hp, 230 V, and speed of 435 r/min DC shunt motor has a nominal full load current 862 A. a. Calculate the total losses and ef

Lostsunrise [7]

Answer:

(a) Losses 11760 watt, Efficiency = 94.06 %

(b) Shunt current 10.21 A

(c) Armature resistance = 0.0084 ohm

Armature voltage = 222.97 volt

Explanation:

Output of the dc motor 2500 hp

As 1 hp = 746 watt

So output power

$P=250\times 746=186500W$

Full load current = 862 A

(A) Input of the motor is equal to

$=VI=230\times 862=198260W$

Therefore losses = input - output

= 192860-186500 = 11760 watt

Efficiency = $\eta =\frac{186500}{192860}=0.9406$ = 94.06 %

(b) Power of shunt field is equal to

$P_{shunt}=11760\times \frac{20}{100}=2352watt$

So shunt current $I_{sh}=\frac{2352}{230}=10.21A$

(C) It is given that 50% of the loss is due to armature current it means 50% loss will be due to shunt current

So loss due to shunt current $=11760\times 0.5=5880W$

Shunt current is equal to $I_{sh}=\frac{5880}{230}=25.56A$

Armature current $I_a=862-25.56=836.43A$

Power loss due to armature = 11760-5880 = 5880 W

Therefore $I_a^2R_a=5880$

$836.43^2\times R_a=5880$

$R_a=0.0084 ohm$

Emf is equal to $E=V-I_aR_a$

$E=230-836.43\times 0.0084=222.97V$

3 0

3 years ago

Explain the 11 sections that a typical bill of quantity is divided into​

Explain the 11 sections that a typical bill of quantity is divided into