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3 years ago

What is greatest velocity? How do you get greatest velocity?

2 answers:

5 0

Velocity is equivalent to a specification of an object’s speed and direction of motion (e.g. 60 km/h to the north). How to get is, Choose a point to the left of the extremum and another point just to the right. If acceleration is negative to the left and positive to the right, the point is a minimum velocity. If acceleration is positive to the left and negative to the right, the point is a maximum velocity. There you go.

aksik [14]3 years ago

4 0

Answer:

Maximum velocity is reached when you stop accelerating, because this is when you can't gain anymore speed, i.e. acceleration is zero. In other words the derivative of velocity is equal to zero. However, zero acceleration can also result in minimum velocity because you can't lose any more speed.

Explanation:

hope this helps

You might be interested in

What property did mosely use to organize the periodic table?

zhenek [66]

Answer:

nitrogen i think but not sure

Explanation:

4 0

3 years ago

What volume of O2(

irakobra [83]

Answer is: volume of oxygen is 4.63 liters.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
<span>R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.</span>
V(O₂) = n·R·T / p.<span>
V(O₂) = 0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.</span><span>
V(O₂<span>) = 4.63 L.</span></span>

4 0

3 years ago

At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution

IRISSAK [1]

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = $x_{A}\times P_{A}^{0}$

vapour pressure of a component (B) in solution = $x_{B}\times P_{B}^{0}$

Where $x_{A},x_{B}$ are mole fraction of component A and B in solution respectively

$P_{A}^{0},P_{B}^{0}$ are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = $0.340\times 745 torr$

= 253 torr

7 0

3 years ago

Calculate the mole fraction of kbr (molar mass 119.00 g/mol) in a solution made by dissolving 0.30 g kbr in 0.400 l of H2O (d =

julia-pushkina [17]

The mole fraction of KBr in the solution is 0.0001

<h3>How to determine the mole of water</h3>

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
Density of water = 1 g/mL
Mass of water =?

Density = mass / volume

1 = Mass of water / 400

Croiss multiply

Mass of water = 1 × 400

Mass of water = 400 g

Finally, we shall determine the mole of the water

Mass of water = 400 g
Molar mass of water = 18.02 g/mol
Mole of water = ?

Mole = mass / molar mass

Mole of water = 400 / 18.02

Mole of water = 22.2 moles

<h3>How to de terminethe mole of KBr</h3>

Mass of KBr = 0.3 g
Molar mass of KBr = 119 g/mol
Mole of KBr = ?

Mole = mass / molar mass

Mole of KBr = 0.3 / 119

Mole of KBr = 0.0025 mole

<h3>How to determine the mole fraction of KBr</h3>

Mole of KBr = 0.0025 mole
Mole of water = 22.2 moles
Total mole = 0.0025 + 22.2 = 22.2025 moles
Mole fraction of KBr =?

Mole fraction = mole / total mole

Mole fraction of KBr = 0.0025 / 22.2025

Mole fraction of KBr = 0.0001

Learn more about mole fraction:

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6 0

2 years ago

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

Andrew [12]

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.

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<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.

6 0

3 years ago