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olga nikolaevna [1]
4 years ago
13

Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s

B→+C→ has zero x and y components.
Engineering
1 answer:
ioda4 years ago
5 0

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

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Explanation:

5 0
3 years ago
Which type of muscle tissue is both voluntary and striated?
Elena L [17]

Answer:

Skeletal muscle

Explanation:

These classifications describe three distinct muscle types: skeletal, cardiac and smooth. Skeletal muscle is voluntary and striated, cardiac muscle is involuntary and striated, and smooth muscle is involuntary and non-striated.

4 0
3 years ago
Read 2 more answers
A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

3 0
4 years ago
Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i
RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

T_0 = T+\frac{V^2}{2c_p}

Where

T = Static temperature

V = Velocity of Fluid

c_p = Specific Heat

Re-arrange to find the static temperature we have that

T = T_0 - \frac{V^2}{2c_p}

T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})

T = 672.88K

Now the pressure of helium by using the Adiabatic pressure temperature is

P = P_0 (\frac{T}{T_0})^{k/(k-1)}

Where,

P_0= Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}

P = 0.399Mpa

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0
3 years ago
thì nghiệm nén xác định cường độ của bê tông trên ba mẫu thí nghiệm hình trụ HxD=300x150(mm). kết quả thu được lực phá hoại P1=4
adell [148]

spanish

Explanation:

the above question is written in spanish

5 0
3 years ago
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