Chemistry, 10.10.2019 05:20 granniejo

Aball is equipped with a speedometer and launched straight upward. the speedometer reading four seconds after launch is shown at the right; the ball is moving downward. at what approximate times would the ball be moving upward and display the following speedometer readings?

Aball is equipped with a speedometer and launched straight upward. the speedometer reading four seco

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Answer from: CeddHendrix

The time shown by the speedometer corresponding to speed of 10 m/s is 1 s.

The time shown by the speedometer corresponding to speed of 20 m/s is 1 s.

The time shown by the speedometer corresponding to speed of 30 m/s is 1 s.

Given data:

The time read by the speedometer is, t = 4 s.

Since the speedometer has low precision, then value approximation can be done.

For the speedometer showing the speed 20 m/s. The time is calculated as,

$v = u +gt_{1}\\$

Here, u is the initial speed and g is the gravitational acceleration and its approximate value is, $g \approx 10 \;\rm m/s^{2}$ .

Solving as,

$10 = 0 +10t_{1}\\t_{1}=1 \;\rm s$

Thus, the time shown by the speedometer corresponding to speed of 10 m/s is 1 s.

For the speedometer showing the speed 20 m/s. The time is calculated as,

$v = u +gt_{2}$

Solving as,

$20 = 10 +10t_{2}\\t_{2}=1 \;\rm s$

Thus, the time shown by the speedometer corresponding to speed of 20 m/s is 1 s.

For the speedometer showing the speed 30 m/s. The time is calculated as,

$v = u +gt_{3}$

Solving as,

$30 = 20 +10t_{3}\\t_{3}=1 \;\rm s$

Thus, the time shown by the speedometer corresponding to speed of 30 m/s is 1 s.

Learn more about first kinematic equation of motion here:

link

Answer from: lilpeepxliltracy

Question 1: 1 s after the motion starts

Question 2: 0 (just when the motion starts)

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

1. From the speedometer shown at the right.

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

Free fall equation: Vf = Vo - gtVo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

2. Time when the speedometer displays a reading of 20 m/s

First, calculate the launching speed:

Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

0 = Vo - gt

Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

20 m/s = 30 m/s - 10 m/s² × tt = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

3. Time when the speedometer displays a reading of 30 m/s

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

Answer from: Quest

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃ → 3 nicl₂ + 2 albarium and copper bromideba + cubr → no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn → znso₄ + na

Answer from: Quest

Sorry i think the answer is b

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