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Ira Lisetskai [31]
3 years ago
7

A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a

t the right; the ball is moving downward. At what approximate times would the ball be moving upward and display the following speedometer readings?

Chemistry
1 answer:
Andrew [12]3 years ago
6 0

Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

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.

<u>1. From the speedometer shown at the right.</u>

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 - gt

  • Vo = 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.

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

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 × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

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

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.

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The average propane cylinder for a residential grill holds approximately 18 kg of propane. how much energy (in kj) is released b
Angelina_Jolie [31]
Let's begin with the basic values  that will be used in the solution.

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Molar Mass Calculation is done as like that
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Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (
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4 years ago
For the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression
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Explanation:

An expression that depicts the ratio of products and reactants raised to the power of their coefficients at equilibrium is called equilibrium constant.

An equilibrium constant is denoted by the symbol 'K'.

For example, the dissociation of nitrous acid in aqueous solution is as follows.

HNO_{2} \rightleftharpoons H^{+} + NO^{-}_{2}

Hence, its expression for equilibrium constant is as follows.

K = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}

Thus, we can conclude that for the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression is K = [ H+] [NO2-] / [HNO2].

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MrMuchimi

Answer:

Initial temperature of the gas cylinder was 167.7K

Explanation:

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<em />

Replacing:

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<h3>Initial temperature of the gas cylinder was 167.7K</h3>
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