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fomenos
3 years ago
14

What are flip flops and what do they look like

Engineering
2 answers:
kondor19780726 [428]3 years ago
8 0
They are what you put in a ground to grow and it turns into a huge tree with flip flops hanging off
Vinil7 [7]3 years ago
4 0
flip flops are shoes and they look like shoes
You might be interested in
Why is the contractor normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract?
just olya [345]

Answer:

It serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

Explanation:

A bid bond is a type of construction bond that protects the obligee in a  construction bidding process.

A bid bond typically involves three parties:

The obligee; the owner or developer of the construction project under bid. The principal; the bidder or proposed contractor.

The surety; the agency that issues the bid bond to the principal example insurance company or bank.

A bid bond generally serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

3 0
3 years ago
Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0
3 years ago
In a Series circuit with two identical loads, the voltage across each load will be:
mel-nik [20]
<h3>Option A</h3>

In a Series circuit with two identical loads, the voltage across each load will be:  the same

<h3><u>Explanation:</u></h3>

A series circuit is one with total the loads in a row. There is barely ONE path for electricity to pass. If this circuit was a series of flashbulbs, and one left out, the left bulbs would switch off. T

he current in a series circuit is universally the same and the voltage over the circuit is the amount of the unique voltage drops over each component. The voltage referred to as a series circuit is equivalent to the amount of the individual voltage drops.

8 0
3 years ago
A rigid canister with a radius of 5 in and a height of 10 in is filled with air. The initial pressure and temperature of air in
Elza [17]

Answer:1.458 Btu

Explanation:

Given

radius of canister\left ( r\right )=5in

Height of canister\left ( h\right )=10 in.

Initial pressure\left ( P_i\right )=14.7 Psi

Initia ltemprature\left ( T_i\right )=70^{\circ}F

Final pressure\left ( P_f\right )=30Psi

as canister is rigid therefore change in volume is zero

therefore

\frac{P_i}{T_i}=\frac{P_f}{T_f}

\frac{14.7}{70}=\frac{30}{T_f}

T_f=142.85^{\circ}F

volume of canister=\pi \times r^{2}\times h

                               =\pi \times 5^{2}\times 10=250\pi

volume of canister=12872,038.8 mm^3

now calculating mass of air

PV=mRT

substituting values

\left ( 14.7Psi\right )\left ( 12872,038.8 mm^3\right )=m\left ( 0.287\right )\left ( 70^{\circ}F\right )

m=21.0192 gm

Therefore heat transferred =mc_p\left ( T_f-T_i\right )

                                             =21.0192\times 10^{-3}\times \left ( 142.857-70\right )

                                             =1539.052J=1.458Btu

3 0
4 years ago
What differentiates an athletic trainer from a clinical exercise physiologist?
kozerog [31]

Answer:

It's number 3

Explanation:

3 0
3 years ago
Read 2 more answers
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