$$A=100\;ft^2 A=200\;ft^2$$ $$C=\$3000$$ $$n=0.6$$ $$x=3000\left(\frac{200}{100}\right)^{0.6}$$ $$=4547.1497\;;\;1980\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;A=200ft$$ $$100-2000;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;n=0.81$$ $$A=1000ft^2$$ $$\frac{C_02000}{C_11980}=\frac{981}{780}\times\left(\frac{1000}{200}\right)^{0.81}$$ $$=\frac{981}{780}\times\left(\frac{1000}{200}\right)^{0.81}\times4547.1497$$ $$=21061.0$$
It is desired to borrow $2000 to meet a financial obligation. This money can be borrowed from a loan agency at a monthly interest rate of 4 percent. Determine the following:(a)The total amount of principal plus simple interest due after 2 years if nointermediate payments are made,(b)The total amount of principal plus compounded interest sue after 2 years if no intermediate payments are made,(c)The nominal interest rate when the interest is compounded monthly,(d)The effective interest rate when the interest is compounded monthly.
(a)the total amountof principal plus simple interest due after 2 years if nointermediate payments are made : P=$2000 R=4% T=2×12=24 $$Simple\;Interest=P\left(1+\frac{r\times t}{100}\right)$$ $$=2000\left(1+\frac{4\times24}{100}\right)$$ =3920 (b)The total amount of principal plus compounded interest due after 2 years if no intermediate payments are made: $$=P\left(1+\frac r{100}\right)^t$$ $$=2000\left(1+\frac4{100}\right)^{24}$$ =5126 (c)The nominal interest rate when the interest rate is…
Water contaminated with small amounts of hazardous waste is gravity-drained out of a large storage tank through a straight commercial steel pipe, 100 mm ID (internal diameter). The pipe is 100 m long with a gate valve near the tank. The entire pipe assembly is mostly horizontal. If the liquid level in the tank is 5.8 m above the pipe outlet, and the pipe is accidently severed 33 m from the tank,compute the flow rate of material escaping from the pipe.
we need some assumptions before solving this question: applying mechanical energy balance between point 1 and point 2 we get $$\frac g{g_c}\bigtriangleup z+F=0$$ for water $$ยต=1.0\times10^{-3}kg/ms$$ $$\rho=1000kg/m^3$$ For pipe entrance K factor , $$K_f=\frac{160}{Re}+0.50$$ For gate valve, $$K_f=\frac{300}{Re}+0.10$$ For pipe exit, $$K_f=1.0$$ for pipe length, $$K_f=\frac{4fL}d=\frac{4f\times33}{0.10}=1320f$$ adding all K factors we get $$\sum K_f=\frac{460}{Re}+1320f+1.60$$ for…
Determine the TLV for a uniform mixture of dusts containing the following particles:Nonasbestiform talc and Quartz with concentration 70,30 and TLV 20,2.7 respectively
given : Type of dust Concentration(wt.%) TLV(mppcf) Nonasbestiform talc 70 20 Quartz 30 2.7 we will use the following formula : $$TLV=\frac1{{\displaystyle\frac{C_1}{TLV_1}}+{\displaystyle\frac{C_2}{TLV_2}}}$$ $$TLV=\frac1{{\displaystyle\frac{0.70}{20}}+{\displaystyle\frac{0.30}{2.7}}}$$ =6.8
Determine the 8hr TWA worker exposure if the worker is exposed to toulene vapours as follows:duration of exposure are 3,3,4 and measured concentration are 100,250,90 respectively.TLV for toulene is 100ppm.is the worker overexposed ?
the following shows simple method to solve this question given: Duration of Exposure(hr) Measured Concentration(ppm) 3 100 3 250 4 90 we can use the equation $$\frac{C_1T_1+C_2T_2+C_3T_3}{T_1+T_2+T_3}$$ on putting the values we get $$\frac{100\ast3+250\ast3+90\ast4}{3+3+4}$$ =141