All categories

4 years ago

Are you able to text without looking at your phone?

1 answer:

5 0

Yes.
Yes
Yes, I told them that it was dangerous.
No.
Both are bad but I think driving drunk is way worse. Yes both take your attention away from the road but driving drunk takes your attention away permanently while texting and driving only takes it away for a few minutes.

You might be interested in

You are given a rectangular piece of cloth with dimensions X by Y, where X and Y are positive integers, and a list of n products

Bond [772]

The proof that recursion is exponential and that dynamic programming is polynomial is given by the formula;

P(x,y) = max{

P(x,y)

max (1 <= h <= X) { P[h, Y] + P(X - h, Y) }

max (1 <= v <= Y) { P[X, v] + P[X, Y - v] }

}

To prove that the recursion is exponential and that dynamic programming is polynomial. we will do so as follows;

Let us first have the assumption that the cloth is in such a manner that either way, a product can be oriented. This implies that that after a cut, we will now have two pieces of cloth.

Now, we will make a list of the side lengths of the products that can fit in the piece after which we will consider a vertical cut for each of the side length as well as a horizontal cut for each of the side length, then we apply the same algorithm to each of the two resulting pieces.

Thus, after the point above, it is likely true that in some instances, there may be a place to cut that is not at a product side length. However, It might be better for us to make a list of lengths composed of one or more pieces side by side as long as the sum is less than the length of the side being considered.

Lastly, we would note that this recursive approach is not limited to just two -dimensional problems as It could also be applied to a single or more than two dimensions. A useful proof would be to prove it for one dimension, then assuming it is true for n dimensions, prove it is true for n + 1 dimensions.

Thus;

P(x,y) = max{

P(x,y)

max (1 <= h <= X) { P[h, Y] + P(X - h, Y) }

max (1 <= v <= Y) { P[X, v] + P[X, Y - v] }

}

Read more at; brainly.com/question/11665190

3 0

3 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

$= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}$

$= 1.751 \times 10^{-5} \% per hr$

4 0

3 years ago

Based on bonding theory, explain why coefficient of expansion increases when you consider ceramics, metals and polymers

Mrrafil [7]

Answer:

The ratio that a material expands in accordance with changes in temperature is called the coefficient of thermal expansion. Because Fine Ceramics possess low coefficients of thermal expansion, their distortion values, with respect to changes in temperature, are low. The coefficients of thermal expansion depend on the bond strength between the atoms that make up the materials.

Explanation:

3 0

4 years ago

Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain

fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

$N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })$