Physics, 01.12.2020 01:00 alexisnichols2001

A block slides across a flat, horizontal surface to the right. For each choice, the arrows represent velocity vectors of the block at successive intervals of time. Which of the following diagrams represents the situation in which the block loses kinetic energy?

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Answer from: aniacopenowell7

The diagrams the question refers to is in the attachment.

Third diagram or alternative C.

Explanation: In physics, vector is a representation of magnitude and direction.

Kinetic Energy is energy made by a moving object, i.e., when an object with mass has velocity, it also has Kinetic Energy. It's calculated as:

$K=\frac{1}{2}mv^{2}$

Vectors below are representing change in velocity during intervals of time.

In the first diagram, vectors doesn't change, so velocity and kinetic energy are constant.

At the second, velocity still constant, however, the "intensity" of the arrow is weaker than the first, which means velocity is less than the first diagram and so is its kinetic energy.

In the fourth diagram, the arrows grow bigger from left to right. This means velocity increases during time. If velocity grows bigger, so does kinetic energy.

In the third diagram, the arrows decrease as it goes to the right: the block's velocity decreases during the successive intervals, which means, kinetic energy diminishes through the interval.

Thus, diagram C represents the situation in which the block loses kinetic energy.

A block slides across a flat, horizontal surface to the right. For each choice, the arrows represent

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

$v = 5.4 t$

after 28s,

$v(28) = 5.4(28) = 151.2$ m/s

which i would round to 150 m/s as my final answer

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