What's the real power if Vrms = 100V, Irms = 2A, and the circuit has a power factor of 0.8?

Since the application can be the prospective employers first impression, it is important to a. always answer questions truthfull

loris [4]

Answer:

Definitely D

Explanation:

You should do all of those things in an application.

7 0

4 years ago

Read 2 more answers

Which principle of the software engineering code of ethics has gilbert violated?.

Yuki888 [10]

Answer:

Judgement

Explanation:

Gilbert is required by the Judgement Principle to "disclose those conflicts of interest that cannot reasonably be avoided or escaped." Since Gilbert professionally believes that the software meets specifications, secures documents, and satisfies user requirements, it is not clear if he violated any principle. However, he could have informed his client of his interest in the software and also presented other software packages of different companies from which the client could make its independent choice.

7 0

3 years ago

The following figures were obtained in a standard tensile test on a specimen of low carbon steel with a circular sectional area:

liq [111]

Answer:

See Explaination

Explanation:

1)here for given stress strain curve graph is given as follows

where for getting stress,S=F/A=4F/(pi*(50*10^-3)^2)

for strain=e=dl/l=dl*10^-3/100 mm/mm or m/m

2)so graph is as follows

3)for getting youngs modulus of elasticity we must know slope of graph stress verses strain and for straight line in elastic region upto 12 point we have elastic region and from that we get E as

E=slope of graph for first 12 points=S/e=14.5665*10^9/.812=17.9390*10^9 N/m2

4)for getting ultimate tensile stress at which specimen bears maximum load without failure so we get UTS as

UTS=maximum load/area=40*10^6/1.9634=20.3728*10^6 N/m2

5)percentage reduction in area is given by

percentage reduction in area=[original area-final area/original area]*100

Percent reduction=[5062-10^2]*100/50^2=96%

6)percentage elongation is given by

percent elongation=[final length-original length/original length]*100

final length at fractureis=14.56+100=114.56 mm

so we get percent elongation as=[114.56-100/100]*100=14.56%

7)true fracture stress is given by load at fracture devided by true area at fracture

Sf=load/(true area)=4*28*10^3/(pi*(10*10^-3)^2)=356.5070*10^6 N/m2

8 0

4 years ago

For a metal that has a yield strength of 690 MPa and a plane strain fracture toughness (KIc) of 32 MPa-m1/2, compute the minimum

Digiron [165]

Answer:

the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

Explanation:

Given the data in the question;

yield strength σ $_y$ = 690 Mpa

plane strain fracture toughness K $_{Ic$ = 32 MPa- $m^{1/2$

minimum component thickness for which the condition of plane strain is valid = ?

Now, for plane strain conditions, the minimum thickness required is expressed as;

t ≥ 2.5( K $_{Ic$ / σ $_y$ )²

so we substitute our values into the formula

t ≥ 2.5( 32 / 690 )²

t ≥ 2.5( 0.0463768 )²

t ≥ 2.5 × 0.0021508

t ≥ 0.005377 m or 5.38 mm

Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

7 0

3 years ago

the hoop is cast on the rough surface such that it has an angular velocity w=4rad/s and an angular acceleration a=5rad/s^2. also

IgorLugansk [536]

Given Information:

Angular velocity = ω = 4 rad/s

Angular acceleration = α = 5 rad/s²

Center deceleration = a₀ = 2 m/s

Required Information:

Acceleration of point A at this instant = ?

Answer:

Acceleration of point A at this instant = 5.94 m/s²

Explanation:

Refer to the attached diagram of the question,

The acceleration of point A is given by

a = a₀ + rα - rω²

Where r is the radial distance between the center and point A, a₀ is the deceleration of center, α is the angular acceleration and ω is the angular velocity.

a = -2i + 0.3j*5k - 0.3j*4²

a = -2i + 1.5(j*k) - 0.3j*16

a = -2i + 1.5(-i) - 4.8j

a = -2i - 1.5i - 4.8j

a = -3.5i - 4.8j

The magnitude of acceleration vector is

a = √(-3.5)² + (-4.8)²

a = √35.29

a = 5.94 m/s²

Therefore, the acceleration of point A is 5.94 m/s²

The angle is given by

θ = tan⁻¹(y/x)

θ = tan⁻¹(-4.8/-3.5)

θ = 53.9°

7 0

4 years ago