OSHA requires safe work practices when working around brake shoes and clutches that contain asbestos.

Income tax payable is included in the account...

den301095 [7]

Answer:

Current Liabilities

Explanation:

Accounts of Current Liabilities

Account Payable
Expense Payable
Tax Outcome
Tax Income
Income tax payable

5 0

2 years ago

Read 2 more answers

Troy must keep track of the amount of refrigerant he uses from a 50-pound cylinder to ensure that accurate

IgorLugansk [536]

Answer:

Amount of gas still in cylinder = 28 pound

Explanation:

Given:

Amount of gas in cylinder = 50 pound

Amount of gas used in Ms. Jones system = 13 pound

Amount of gas used in client system = 9 pound

Find:

Amount of gas still in cylinder

Computation:

Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system

Amount of gas still in cylinder = 50 - 13 - 9

Amount of gas still in cylinder = 28 pound

7 0

3 years ago

From your cooling load (8890.007 Btu/hr = 2.605kW, determine mass flow rate of refrigerants. Use the following "rule of thumb" e

NeX [460]

Answer:

0.740833917 ton/hr

Explanation:

Given:

Cooling load, 8890.007 Btu/hr = 2.605 kW

Room size = 180 $ft^{2}$

According to the thumb rule

1 ton of refrigerant = 12000Btu

Hence for 8890.007 Btu/hr,

the mass flow rate of the refrigerant is =8890.007 / 12000

= 0.740833917 ton per hr

Hence, mass flow rate is 0.740833917 ton/hr

7 0

3 years ago

A steel wire is suspended vertically from its upper end. The wire is 400 ft long and has a diameter of 3/16 in. The unit weight

jek_recluse [69]

Answer:

a) the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb

Explanation:

Given the data in the question;

Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in

Diameter d = 3/16 in

Unit weight w = 490 pcf

First we determine the area of the wire;

A = π/4 × d²

we substitute

A = π/4 × (3/16)²

A = 0.0276 in²

Next we get the Volume

V = Area × Length of wire

we substitute

V = 0.0276 × 4800

V = 132.48 in³

Weight of the steel wire will be;

W = Unit weight × Volume

we substitute

W = 490 × ( 132.48 / 12³ )

W = 490 × 0.076666

W = 37.57 lb

a) the maximum tensile stress due to the weight of the wire;

σ $_w$ = W / A

we substitute

σ $_w$ = 37.57 / 0.0276

= 1361.23 psi

Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi

Maximum load P that the wire can safely support its lower end will be;

P = ( σ $_{all$ - σ $_w$ )A

we substitute

P = ( 24000 - 1361.23 )0.0276

P = 22638.77 × 0.0276

P = 624.83 lb

Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb

5 0

2 years ago

A geothermal pump is used to pump brine whose density is 1050 kg/m3 at a rate of 0.3 m3/s from a depth of 200 m. For a pump effi

nasty-shy [4]

Answer:

Input power of the geothermal power will be 686000 J

Explanation:

We have given density of brine $\rho =1050kg/m^3$

Rate at which brine is pumped $V=0.3m^3/sec$

So mass of the pumped per second

Mass = volume × density = $1050\times 0.3=315$ kg/sec

Acceleration due to gravity $g=9.8m/sec^2$

Depth h = 200 m

So work done $W=mgh=315\times 9.8\times 200=617400J$

Efficiency is given $\eta =0.9$

We have to fond the input power

So input power $=\frac{617400}{0.9}=686000J$

So input power of the geothermal power will be 686000 J

5 0

3 years ago