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Gemiola [76]
2 years ago
6

A. 1.2KΩ = Ωwhat is the ohms

Engineering
1 answer:
mihalych1998 [28]2 years ago
3 0

Answer:

1200

Explanation:

ohms is the si unit for resistance.

hence, 1.2 k(ohms) = 1200 ohms since k = kilo = ×1000

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A ________ is a condition or capability needed by a user to solve a problem or achieve an objective that satisfies a standard or
Aliun [14]
A successful brain to help
7 0
3 years ago
A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto
vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

4 0
4 years ago
SB-4 Which one of the following is true about red buoys under the U.S. Aids to Navigation System?
stich3 [128]

Question:

1. Some are known as "nun" buoys

2. They are labeled with odd numbers

3. If it is lighted, the light color is green

4. Some are known as "can" buoys

Answer:

The correct option is;

1. Some are known as "nun" buoys

Explanation:

Based on the lateral system, on the starboard side, one can find the red even numbered marks while the odd-numbered, green, marks are located on the port side of a channel such that the buoy numbers increase as a vessel travels upstream.

The red buoys are cones shaped in appearance and have triangular reflective sign markings embossed and they are of different types included in the order of lower water depth

1. NUN buoy

2. Lighted buoy

3. Light

4. Day beacon

Therefore, the correct option is 1. Some are known as "NUN" buoys.

8 0
4 years ago
A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac
Alex Ar [27]

Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

 ⇒d_2=28.60 mm

So throat diameter d_2=28.60 mm

     

4 0
3 years ago
Con que otro nombre se le conoce a los delitos informaticos
kow [346]
Repuesto: Cyberdelito
7 0
3 years ago
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