Best budget freestyle drone?

What is the force in kN of work done is 1.2 ms moves through 120m

Semmy [17]

Answer:

$\frac{1.2}{120}$

0.01

5 0

2 years ago

Pam Jones worked for 41 years at the same company and had positive performance ratings and personnel records. She needed a calcu

UNO [17]

Answer:

a) The key issues are the sue for libel and the evidence.

b) I would make a deal with her and implement a security program in the company.

Explanation:

The main issue in this case is that Pam Jones sued the company for libel, and the company remains in a position in which it has to prove that the internal investigation followed the right steps and indeed, the proves reflected that she was guilty and the fact that she got fired was correct.

The important here is exactly that the theft can be proved.

As an HR Director, I would give the correct proves in order to win the case, and if that is impossible, because of the tools and evidence, I would make a deal with her where both parts can be adequate to the problem.

She can´t be working again in the company but she can get financed according to her working years; also I would use this case as a growing opportunity by implementing new security methods that give more confidence between the company and its employees.

5 0

3 years ago

In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26

scZoUnD [109]

Answer:

The strength coefficient is $625$ and the strain-hardening exponent is $0.435$

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find $(K)$ and the $(n)$ .

$\sigma =K\epsilon^n$

We will plug the values in the formula.

$250=K\times (0.12)^n\\350=K\times (0.26)^n$

We will solve these equation.

$K=\frac{250}{(0.12)^n}$ plug this value in $350=K\times (0.26)^n$

$350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n$

Taking a natural log both sides we get.

$ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435$

Now, we will find value of $K$

$K=\frac{250}{(0.12)^n}$

$K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625$

So, the strength coefficient is $625$ and the strain-hardening exponent is $0.435$ .

5 0

3 years ago

A 4-pole, 60-Hz, 690-V, delta-connected, three-phase induction motor develops 20 HP at full-load slip of 4%. 1) Determine the to

gladu [14]

Answer:

1. i. 20 Nm ii. 4.85 HP

2. 16.5 %

Explanation:

1) Determine the torque and the power developed at 4% slip when a reduced voltage of 340V is applied.

i. Torque

Since slip is constant at 4 %,torque, T ∝ V² where V = voltage

Now, T₂/T₁ = V₂²/V₁² where T₁ = torque at 690 V = P/2πN where P = power = 20 HP = 20 × 746 W = 14920 W, N = rotor speed = N'(1 - s) where s = slip = 4% = 0.04 and N' = synchronous speed = 120f/p where f = frequency = 60 Hz and p = number of poles = 4.

So, N' = 120 × 60/4 = 30 × 60 = 1800 rpm

So, N = N'(1 - s) = 1800 rpm(1 - 0.04) = 1800 rpm(0.96) = 1728 rpm = 1728/60 = 28.8 rps

So, T = P/2πN = 14920 W/(2π × 28.8rps) = 14920 W/180.96 = 82.45 Nm

T₂ = torque at 340 V, V₁ = 690 V and V₂ = 340 V

So, T₂/T₁ = V₂²/V₁²

T₂ = (V₂²/V₁²)T₁

T₂ = (V₂/V₁)²T₁

T₂ = (340 V/690 V)²82.45 Nm

T₂ = (0.4928)²82.45 Nm

T₂ = (0.2428)82.45 Nm

T₂ = 20.02 Nm

T₂ ≅ 20 Nm

ii. Power

P = 2πT₂N'

= 2π × 20 Nm × 28.8 rps

= 1152π W

= 3619.11 W

converting to HP

= 3619.11 W/746 W

= 4.85 HP

2) What must be the new slip for the motor to develop the same torque when the reduced voltage is applied

Since torque T ∝ sV² where s = slip and V = voltage,

T₂/T₁ = s₂V₂²/s₁V₁²

where T₁ = torque at slip, s₁ = 4% and voltage V₁ = 690 V and T₂ = torque at slip, s₂ = unknown and voltage V₂ = 340 V

If the torque is the same, T₁ = T₂ ⇒ T₂T₁ = 1

So,

T₂/T₁ = s₂V₂²/s₁V₁²

1 = s₂V₂²/s₁V₁²

s₂V₂² = s₁V₁²

s₂ = s₁V₁²/V₂²

s₂ = s₁(V₁/V₂)²

substituting the values of the variables into the equation, we have

s₂ = s₁(V₁/V₂)²

s₂ = 4%(690/340)²

s₂ = 4%(2.0294)²

s₂ = 4%(4.119)

s₂ = 16.47 %

s₂ ≅ 16.5 %

3 0

3 years ago

Electricians will sometimes call _____ “disconnects” or a “disconnecting means”.

AleksandrR [38]

Answer:
D) Three-pronged plug

Hope it helps u!

4 0

3 years ago