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dimulka [17.4K]
1 year ago
6

Hwan also wants to determine the depreciation balance for the first year and the last year of the useful life of the product har

dware. Determine these amounts as follows: In cell E23, enter a formula using the SYD function to calculate the depreciation balance for the first year. b. Use Year 1 (cell C18) as the current period. In cell E24, enter a formula using the SYD function to calculate the depreciation balance for the last year. d. Use Year 6 (cell H18) as the current period.
Engineering
1 answer:
Elden [556K]1 year ago
4 0

In Menlo Park, California, Hwan Rhee is debating whether to launch ChargeAll, a software firm that will create full-room wireless chargers for any kind of mobile electronic device.

In Menlo Park, California, Hwan Rhee is debating whether to launch ChargeAll, a software firm that will create full-room wireless chargers for any kind of mobile electronic device. In order to examine the financial information for a startup loan that would pay for the components and product manufacture, Hwan is using an Excel worksheet. He requests your assistance in fixing mistakes and performing financial computations in the worksheet.

Visit the worksheet for loan analysis. Hwan requests that you fix the mistakes in the worksheet before he can compute the principal and interest payments on the loan.

To know more about software click here:

brainly.com/question/985406

#SPJ4

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Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a
Elena L [17]

Answer:

\mathbf{\tau_c =5.675 \ MPa}

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [\bar 101]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]

where;

[d_1\ e_1 \ f_1] = directional indices for tensile stress

[d_2 \ e_2 \ f_2] = slip direction

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_2 = -1 , e_2 = 0 , f_2 = 1

cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]

cos \ \lambda = \dfrac{1}{\sqrt{2}}

Also, to find the angle \phi between the stress [001] & normal slip plane [111]

Then;

cos \  \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_3 = 1 , e_3 = 1 , f_3 = 1

cos \  \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]

cos \phi= \dfrac{1} {\sqrt{3} }

However, the critical resolved SS(shear stress) \mathbf{\tau_c} can be computed using the formula:

\tau_c = (\sigma )(cos  \phi )(cos \lambda)

where;

applied tensile stress \sigma = 13.9 MPa

∴

\tau_c =13.9\times (  \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})

\mathbf{\tau_c =5.675 \ MPa}

3 0
3 years ago
What is the volume of the rectangular prism shown.
Likurg_2 [28]
It would be 72cm bc u need to add up all the line in the back to
7 0
3 years ago
Sinks must be used for the correct intended purpose to prevent
irakobra [83]

Answer:

... spilling water or getting anything cascading onto the floor

8 0
2 years ago
Tires can be recycled instead of thrown out.<br> True<br> False
Arisa [49]

Answer:

True :)

Explanation:

You can recycle it! Tire recycling is the most practical and environment-friendly way of disposing of old and worn-out tires. Due to their inherent durability, large volume and environment and health risks, tires are one of the most problematic sources of solid wastes.

Hope it helped have a nice day! :)

8 0
2 years ago
A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi
raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation =\frac{N_1 +N_2}{2}

Nm = \frac{500+750}{2} = 625 rpm

w =\frac{2\pi Nm}{60}

  =65.44 rad/sec

F_T = mw^2 e

F_T = mew^2

       = 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio =\frac{300}{428.36} = 0.7

also

transmission ratio = \frac{1}{[\frac{w}{w_n}]^{2} -1}

0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}

SOLVING FOR Wn

Wn = 42 rad/sec

Wn = \sqrt {\frac{k}{m}

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

3 0
3 years ago
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