Metal and dirt are not considered contaminants to oll. A) O True B) O False

Flip-flops are normally used for all of the following applications, except ________. logic gates data storage frequency division

SpyIntel [72]

Flip-flops are normally used for all of the following applications, except logic gates.

<h3>What are Flip flops?</h3>

Flip flops are known to be tools that are used for counting. They come in different ranges.

Note that Flip flops are one that can be seen on counters, storage registers, and others and as such, Flip-flops are normally used for all of the following applications, except logic gates.

Learn more about Flip flops from

brainly.com/question/4237777

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6 0

2 years ago

Define Viscosity. What are the main differences between viscous and inviscid flows?

Evgesh-ka [11]

1. Define Viscosity

In physics, Viscosity refers to the level of resistance of a fluid to flow due to internal friction, in other words, viscosity is the result of the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow. For example, the honey is a fluid with high viscosity while the water has low viscosity.

What are the main differences between viscous and inviscid flows?

Viscous flows are flows that has a thick, sticky consistency between solid and liquid, contain and conduct heat, does not have a rest frame mass density and whose motion at a fixed point always remains constant. Inviscid flows, on the other hand, are flows characterized for having zero viscosity (it does not have a thick, sticky consistency), for not containing or conducting heat, for the lack of steady flow and for having a rest frame mass density

Furthermore, viscous flows are much more common than inviscid flows, while this latter is often considered an idealized model since helium is the only fluid that can become inviscid.

5 0

3 years ago

A teenager was pulling a prank and placed a large stuffed penguin in the middle of a roadway. A driver is traveling on this leve

Anvisha [2.4K]

Whats the question????

3 0

3 years ago

A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at

Radda [10]

Answer:

0.59

Explanation:

The batch is rejected if any of the random samples are found defective, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

$\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}$

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, the probability that the batch is rejected is 1 - 0.41 = 0.59.

4 0

3 years ago

Here, we want to become proficient at changing units so that we can perform calculations as needed. The basic heat transfer equa

netineya [11]

Answer:

9500 kJ; 9000 Btu

Explanation:

Data:

m = 100 lb

T₁ = 25 °C

T₂ = 75 °C

Calculations:

1. Energy in kilojoules

ΔT = 75 °C - 25 °C = 50 °C = 50 K

$m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}$

2. Energy in British thermal units

$\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}$

7 0

3 years ago