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IrinaVladis [17]
3 years ago
6

A set of bus bars also can be called a

Engineering
2 answers:
const2013 [10]3 years ago
8 0
A bus bar is also known as a panel board
Vikki [24]3 years ago
3 0

Answer:

hi

Explanation:

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What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP
ella [17]

Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

Yield strength = 500 MPa

Strain = 0.003

Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

3 0
4 years ago
Read 2 more answers
A lightweight electric car is powered by ten 12-V batteries. At a speed of 100 km/h, the average frictional force is 1500 N. Wha
Aleksandr [31]

Answer:41.67kW

Explanation: a) power = force * velocity

Convert km/hr to m/s=1000m/(60×60s)

=1000/3600

P=F×vel

Power =1500×100× 1000/3600

=1500×100×5/18

=41,666.67W

=41.67kW

6 0
3 years ago
A meter stick can be read to the nearest millimeter and a travelling microscope can be read to the nearest 0.1 mm. Suppose you w
german

Answer: No

Explanation:

Length= 2cm= 20mm

Now meter stick can read to nearest millimeter.

It is given that length is to be measured with a precision of 1% of 20mm= 1/100 * 20= 0.2mm

Since the least count is 1mm of meter stick and precision required is less than that. So, meter stick cannot be used for this, travelling microscope can be used for this as it can read to 0.1mm.

3 0
3 years ago
What must engineers keep in mind so that their solutions will be appropriate?
vekshin1

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

7 0
3 years ago
A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its
ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

F = 23.33 kN

Applying the impulse momentum principle on the three cars;

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F(75-0) = 1.125 × 10⁶

The force T developed is then;

T =  1.125 × 10⁶ / 75

T = 15000 N

T = 15 kN

7 0
3 years ago
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