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elena55 [62]
3 years ago
12

As of December 31, 2013, Stand Still Industries had $2,500 of raw materialsinventory. At the beginning of 2013, there was $2,000

of materials on hand. During theyear, the company purchased $325,000 of materials; however, it paid for only $312,500.How much inventory was requisitioned for use on jobs during 2013?
Physics
1 answer:
galben [10]3 years ago
5 0

Answer:

$ 324,500

Explanation:

given,

stand still industry raw material inventory = $2,500

material in hand at the beginning of 2013 = $ 2000

country purchases the material of $325,000

country paid only = $312,500

inventory use in 2013 = ?

inventory requisitioned for use on job 2013 is

                         =  $ 325,000 + $ 2,000 - $ 2,500                                

                         = $ 324,500                                        

hence, the inventory requisitioned for use on job 2013 is equal to  $ 324,500

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Tool
Ede4ka [16]

Answer:

9758 how many significant figures

3 0
2 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
If you traveled one mile at a speed of 100 miles per hour and another mile at a speed if 1 mile per hour, your average speed wou
Semenov [28]

Answer:

v = 1.98 mph

Explanation:

Given that,

Speed to travel one mile is 100 mph

Speed to travel another mile is 1 mph

The formula used to find your average speed is given by :

v=\dfrac{2v_1v_2}{v_1+v_2}

Putting the values, we get :

v=\dfrac{2\times 100\times 1}{100+1}

v = 1.98 mph

So, yours average speed is 1.98 mph.

7 0
3 years ago
A block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the
bezimeni [28]

Answer:

The answer is "a, c and b"

Explanation:

  • Its total block power is equal to the amount of potential energy and kinetic energy.
  • Because the original block expansion in all situations will be the same, its potential power in all cases is the same.
  • Because the block in the first case has no initial speed, the block has zero film energy.
  • For both the second example, it also has the v_o velocity, but the kinetic energy is higher among the three because its potential and kinetic energy are higher.
  • While over the last case the kinetic speed is greater and lower than in the first case, the total energy is also higher than the first lower than that of the second.
  • The greater the amplitude was its greater the total energy, therefore lower the second, during the first case the higher the amplitude.
4 0
2 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
Read 2 more answers
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