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3 years ago

Please need help on this

Physics

1 answer:

Vera_Pavlovna [14]3 years ago

4 0

Answer:

i have no idea

Explanation:

it could be B. but i dont belive in myself sooooooooooo.....ehhhh yeah

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Help !!! Pick all the apply

Sergio [31]

The answer is the Second one

6 0

3 years ago

Please help :)<br> i would rlly appreciate it :)))

algol [13]

Answer:

the answer for that is "A"

6 0

3 years ago

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

3 0

3 years ago

The force F shown in Figure 4.30 has a moment of 40 Nm about the pivot. Calculate the magnitude

salantis [7]

$\boxed{\sf \tau=rFsin\theta}$

Put values

$\\ \rm\hookrightarrow 40=2Fsin40$

$\\ \rm\hookrightarrow Fsin40=20$

$\\ \rm\hookrightarrow 0.64F=20$

$\\ \rm\hookrightarrow F=31.25N$

8 0

2 years ago

5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur

Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

Л = 1 - $\frac{Tl}{Th}$

where Tl = 300 k

Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 ) :

= P3v3 = mRT3 ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

= (1076.25 * 0.4 ) / 1237.2

= 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

= 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below

6 0

3 years ago