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Ilya [14]
2 years ago
6

Oil with a specific gravity of 0.72 is used as the indicating fluid in a manometer. If the differential pressure across the ends

of the manometer is 6kPa, what will be the difference in oil levels in the manometer?
Engineering
1 answer:
Elan Coil [88]2 years ago
6 0

Answer:

the difference in oil levels is 0.850 m

Explanation:

given data

specific gravity ρ = 0.72

pressure across P = 6 kPa = 6000 Pa

solution

we get here difference in oil levels h is

P = ρ × g × h   .................1

here ρ = 0.72 × 1000 = 720 kg/m³

and g is 9.8  

put here value in equation 1  and we get h

6000 = 720  × 9.8 × h

h = \frac{6000}{720\times 9.8}  

h = 0.850 m

so the difference in oil levels is 0.850 m

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A flexural member is fabricated from two flange plates 7-1/2 x ½ and a web plate 17 x 3/8. The yield stress of steel is 50 ksi.
MissTica

Answer:

(a)

Plastic section modulus Z =92.72 in^3

Plastic moment Mp = 4636 kip-in

(b)

Elastic section modulus S = 80.88 in^3

Yield moment My = 4044 kip-in

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached files.i hope my explanation will help you in understanding this particular question.

7 0
2 years ago
(a) If you needed to fit an acrylic base in a box that is 250mm x 250mm square, and the kerf on the laser cutter is 0.3mm, what
Kazeer [188]

Answer:

l = 250.3\,mm

Explanation:

The size needed to use the kerf on the laser properly is:

l = 250\,mm + 0.3\,mm

l = 250.3\,mm

3 0
3 years ago
The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m/s and its acceleration is 7 m/s 2 .
Neporo4naja [7]

Answer:

Velocity components

V_r = -16.28 m/s

V_z = -22.8 m/s

V_q = 0 m/s

For Acceleration components;

a_r = -4.07m/s^2

a_z = -5.70m/s^2

a_q = 0m/s^2

Explanation:

We are given:

Speed v_o = 28 m/s

Acceleration a_o= 7 m/s^2

We first need to find the radial position r of washer in x-y plane.

Therefore

r = \sqrt{300^2 + 400^2}

r = 500 mm

To find length along direction OA we have:

L = \sqrt{500^2 + 700^2}L = 860 mm

Therefore, the radial and vertical components of velocity will be given as:

V_r = V_o*cos(Q)

V_z = V_o*sin(Q)

Where Q is the angle between OA and vector r.

Therefore,

V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}

V_r = -16.28 m/s

• V_z = 28 * \frac{700}{860} = -22.8

• V_q = 0 m/s

The radial and vertical components of acceleration will be:

a_r = a_o*cos(Q)

a_z = a_o*sin(Q)

Therefore we have:

• a_r = 7* \frac{500}{860} = -4.07m/s^2

• a_z = 7 * \frac{700}{860} = -5.70 m/s^2

• a_q = 0 m/s^2

Note : image is missing, so I attached it

3 0
2 years ago
Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb
Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = \sqrt{400^2 + 650^2 - 2(400)(650)cos30}

      = \sqrt{160000 + 422500 - 80210}

      = 708.72 Ib

using the sine law to determine the design angle

\frac{sin}{400}  = \frac{sin 30}{Fac}

hence ∅ = sin^{-1} (\frac{sin 30 *400}{708.72} )

              = sin^{-1} 0.0871 =  4.9968 ≈ 5⁰

4 0
2 years ago
what is the advantage of decreasing the field current of a separately excited dc motor below its nominal value
enyata [817]

Answer:

Ability to rotate at higher speeds

Explanation:

Constant K1 becomes greater than the other constant K2

This translates to that the motor being able to rotate at high speeds, without necessarily exceeding the nominal armature voltage.

The armature voltage is the voltage that is developed around the terminals of the armature winding of an Alternating Current, i.e AC or a Direct Current, i.e DC machine during the period in which it tries to generate power.

6 0
2 years ago
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