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lianna [129]
4 years ago
8

Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the approp

riate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.

Engineering
1 answer:
Anvisha [2.4K]4 years ago
6 0

Answer:

The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.

Explanation:

The detailed steps and appropriate calculation is as shown in the attached file.

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Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 65 and standard deviati
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Answer:

n this question, we are asked to find the probability that  

R1 is normally distributed with mean 65  and standard deviation 10

R2 is normally distributed with mean 75  and standard deviation 5

Both resistor are connected in series.

We need to find P(R2>R1)

the we can re write as,

P(R2>R1) = P(R2-R1>R1-R1)

P(R2>R1) = P(R2-R1>0)

P(R2>R1) = P(R>0)

Where;

R = R2 - R1

Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.

u = u2-u1

u = 75 - 65 = 10ohm

sd = √sd1² + sd2²

sd = √10²+5²

sd = √100+25 = 11.18ohm

Now we will calculate the z-score, to find  P( R>0 )

Z = ( X -u)/sd

the z score of 0 is

z = 0 - 10/11.18

z= - 0.89

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4 years ago
A rigid bar ABCD is pinned at A and supported by two steel rods connected at B and C, as shown. There is no strain in the vertic
mylen [45]

Answer:

See attached picture.

Explanation:

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3 years ago
A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an
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Answer:

The right solution is "28.45%".

Explanation:

The given values are:

P_4=50\ kPa

h_4=0.7(2304.7)+340.5

    =1953.83 \ KJ/Kg

and,

P_3=15 \ mPa

h_3=hg

    =2610.8 \ KJ/Kg

s_3=sg

    =5.3108 \ KJ/Kgh

At 45,

⇒ x_{45} = \frac{5.3108-1.0912}{6.5019}

          =0.66

At P_4=50 \ Kpa,

h_f=340.54

or,

V_f=0.001030 \ m^3/Kg

then,

⇒ h_2=340.54+0.001030(15\times 10^{3}-50)

        =355.94 \ kJ/kg

hence,

The isentropic efficiency of turbine will be:

⇒ n_T=\frac{h_3-h_4}{h_3-h_{45}}

         =\frac{2610.8-1953.83}{2610.8-1836.26}

         =84.818 (%)

The thermal efficiency of cycle will be:

⇒ n_C=\frac{W_T-W_P}{2_{in}}

         =\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}

         =28.45 (%)  

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3 years ago
You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your
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Solution:

The given formula,

x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}

y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}

\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}

From the table,

1) \(\quad x=2 \cdot 87, \quad y=0.96\)\\\(\frac{x}{y}=\frac{2187}{0.96}\)22895\\\\2) \(x=3 \cdot 466 \cdot y=6 \cdot 998\)\\\(\frac{x}{y}=\frac{3 \cdot 466}{0.898}\)\(=3 \cdot 4729\)

3\(x=3 \cdot 229, y=1 \cdot 08\)\\\(\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}\)\\=2.9898\)\\\\4) \(x=6.525, y=1.094\)\\\(\frac{x}{y}=\frac{5.625}{1.094}\)\\=5.0502

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Answer: Outside an intersections

Explanation:

6 0
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