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3 years ago

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Here u go vagdhf dis day did. Du video ioi Hi I gotta go to do something fun to do something that would you want to me see you l

ater love you bye bye babe love y'all bye love ❤️

1 answer:

dalvyx [7]3 years ago

5 0

Ummmm what? Idk how to answer that

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The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both

DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

L^2 = x_a^2 + y_b^2 ....... 1

y_b = sqrt( 20^2 - 16^2 )

y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

0 = 2*x_a*v_a + 2*y_b*v_b

0 = x_a*v_a + y_b*v_b ..... 2

0 = 16*36 + 12*v_b

v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

0 = 9 + 4*a_a/3 + 16 + a_b

4*a_a/3 + a_b = -25

4*a_a + 3*a_b = -75 .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

F_x = m_a*a_a

P - R_r*16/20 = m_a*a_a

- For Slider B, Apply Newton's second law of motion in y direction:

F_y = m_b*a_b

- R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

P - 4*m_b*a_b / 3 = m_a*a_a

3P = 3*m_a*a_a + 4*m_b*a_b ... 4

- Where, P = Is the force acting on slider A

P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.

5 0

2 years ago

The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat

lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar

Saturation temperature corresponding to saturation pressure =40°C

$h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}$

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

$Total\ heat\ =C_p\Delta T+\Delta h$

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

$h_1=\dfrac{\Delta h}{\Delta T}$

$h_1=\dfrac{2406}{20}$

$h_1=120.3\frac{KJ}{kg-m^2K}$

Now film coefficient after inclusion of sensible heat

$h_2=\dfrac{total\ heat}{\Delta T}$

$h_2=\dfrac{2,544}{20}$

$h_2=127.2\frac{KJ}{kg-m^2K}$

$So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100$

$=\dfrac{127.2-120.3}{120.3}\times 100$

=5.75 %

So Percentage change 5.75 %.

3 0

2 years ago

An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.

saul85 [17]

Answer:

a. $L_o$ = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, $L_o$ = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

$L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)$

For the school, $K_{LL}$ = 2

Therefore, we have;

$L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf$

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, $W_d$ = b × $W_{D + L}$ =

∴ $W_d$ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, $W_d$ = 812.8 ft./lb

d. For the uniformly distributed load, we have;

$V_{max}$ = 812.8 × 26/2 = 10566.4 lbs

$M_{max}$ = 812.8 × 26²/8 = 68,681.6 ft-lbs

$v_{max}$ = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

$P_L$ = 1000 lb

$W_{D}$ = 20 × 16 = 320 lb/ft.

$V_{max}$ = 1,000 + 320 × 26/2 = 5,160 lbs

$M_{max}$ = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

$v_{max}$ = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0

2 years ago

What is the inductive reactance of a 20 mH inductor at a frequency of 100Hz?

Inessa05 [86]

Answer:

The correct approach is "12.56 Ω".

Explanation:

The given values are:

Frequency,

f = 100 Hz

Inductor length,

L = 20 mH

Now,

The inductive reactance will be:

⇒ $X_L=2 \pi fL$

On putting the estimated values, we get

⇒ $=2 (3.14)\times 100\times 20\times 10^{-3}$

⇒ $=12.56 \ \Omega$

8 0

2 years ago

Which option identifies the most likely scaling factor in the above drawing of the Mercury capsule and booster unit? 1:1,000 1:1

Elina [12.6K]

Answer:

I would say 10000;1 but thats just me

Explanation:

3 0

3 years ago