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3 years ago

At sea level, the mass of a watermelon is 5 kg in Sl units. Then the weight of it in English units is: _______.

Engineering

2 answers:

Lapatulllka [165]3 years ago

4 0

Answer:

11.0lbm

Explanation:

1 kg = 2.20 lbm

iogann1982 [59]3 years ago

3 0

Answer:

C. 11.0 lbm

Explanation:

1 lbm = 0.45359 kg

Therefore, 5 kg of watermelon = 5kg × 1 lbm/0.45359kg = 11.0 lbm

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An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat

IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature $T_L$ = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature $T_H$ = 99° F = ( 99 + 460 )R = 559 R

Cooling Load $Q_L$ = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

Coefficient of performance of the air conditioner;

$COP_{air-condition$ = Cooling Load $Q_L$ / power P

we substitute

$COP_{air-condition$ = 30000 Btu/h / 8396.619 Btu/h

$COP_{air-condition$ = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

- Power input required ( in hp )

$Q_L$ / $P_{required$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

30000 Btu/h / $P_{required$ = 530 R / ( 559 R - 530 R )

30000 Btu/h / $P_{required$ = 530 R / 29 R

we solve for $P_{required$

$P_{required$ = ( 30000 Btu/h × 29 R ) / 530 R

$P_{required$ = ( 870000 Btu/h / 530 )

$P_{required$ = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

$P_{required$ = ( 1641.5094 / 2544.43 ) hp

$P_{required$ = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

$COP_{rev-air-condition$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

$COP_{rev-air-condition$ = 530 R / ( 559 R - 530 R )

$COP_{rev-air-condition$ = 530 R / 29 R

$COP_{rev-air-condition$ = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0

3 years ago

Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive mach

STatiana [176]

Answer:

Explained

Explanation:

Cutting tools:

1. Cutting tools can either be single point or multi point.

2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.

3.Cutting tools have definite shapes and geometry.

Abrasive machining tools

1. Abrasive tools are always multi point tools.

2. Abrasive tools composed of abrasives bounded in medium of resin or metal.

3. They do not have definite geometry of shape

7 0

4 years ago

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon

balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P= $2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power= $\frac{work done}{time}$

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ

7 0

4 years ago

Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at

AysviL [449]

Answer:

(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

State1 : p1 = 100kpa, T1= 25+273 =298k

From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

State 2a:p2=2mpa,t2=540k (actual condition 2a)

h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

=246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

=s20-s10-Rln(p2/p1), T0=250C

=(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

= 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

= -38.39 kJ/kg

a)Reversible work = Win actual -any irreversiblities present

=246.17 + irreversibilty

=246.17+-38.39

=207 kJ/kg

8 0

3 years ago

A heat exchanger takes compressed liquid water and converts it into steam with a temperature and pressure of 20 Mpa and 480 C; r

belka [17]

Answer:

See explaination

Explanation:

Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.

Please kindly check attachment for the step by step solution of the given problem.

6 0

3 years ago