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adelina 88 [10]
2 years ago
5

A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did

it take the car to reach this final velocity?
Physics
1 answer:
dezoksy [38]2 years ago
8 0

Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

  • \boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}

It divides:

  • \boxed{t=5\ s}

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

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A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w
barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

Moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}

Moles\ of\ AgBr= 0.1225\ mol

The reaction taking place is:

MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4

From the reaction,

4 moles of AgBr is produced when 1 mole of MBr_4 undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of MBr_4 undergoes reaction

0.1225 mole of AgBr is produced when \frac {1}{4}\times 0.1225 mole of MBr_4 undergoes reaction

Moles of MBr_4 got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

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0.030625 = \frac{12.5843}{319.616 + x}

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<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

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2 years ago
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