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alisha [4.7K]
3 years ago
6

17. Swing arm restraints are intended to prevent a vehicle from falling off a lift.

Engineering
1 answer:
Dimas [21]3 years ago
5 0

This is about usage of Swing arm restraints.

<em><u>B) False</u></em>

There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.

  • Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.

  • This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.

Read more at; brainly.com/question/17972874

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There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.
sergejj [24]

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})

C_{1}=1.2606\times10^{-5} $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})

C_{2}=3.334\times10^{-5} $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})

C_{3}=1.66\times10^{-5} $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

 

5 0
3 years ago
An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th
sp2606 [1]

Answer:

<em> - 14.943 W/m^2K  ( negative sign indicates cooling ) </em>

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

<u>Calculate the overall heat loss coefficient </u>

Note : heat lost by water = heat loss through convection

m*Cp*dT  = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp  ( integrate the relation )

In ( \frac{49-8}{60-8} ) =  h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

              = - 14.943  W/m^2K  ( heat loss coefficient )

7 0
3 years ago
What’s the symbol for elevation
nataly862011 [7]

Answer:

See image

Explanation:

textbook images

4 0
3 years ago
3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue
Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0
3 years ago
Substances A and B have retention times of 16.63 and 17.63 min, respectively, on a 30 cm column. An unretained species passes th
Svet_ta [14]

Answer:

The time required to elute the two species is 53.3727 min

Explanation:

Given data:

tA = retention time of A=16.63 min

tB=retention time of B=17.63 min

WA=peak of A=1.11 min

WB=peak of B=1.21 min

The mathematical expression for the resolution is:

Re_{s} =\frac{2(t_{B}-t_{A})}{W_{A}+W_{B} } =\frac{2*(17.63-16.63)}{1.11+1.21} =0.8621

The mathematical expression for the time to elute the two species is:

\frac{t_{2}}{t_{1}} =(\frac{Re_{B} }{Re_{s} } )^{2}

Here

ReB = 1.5

t_{2} =t_{1} *(\frac{Re_{B} }{Re_{s} } )^{2} =17.63*(\frac{1.5}{0.8621} )^{2} =53.3727min

6 0
3 years ago
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