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fiasKO [112]
2 years ago
10

a vehicle accelerates with 0.4 m/s. calculate tue time taken by the vehicle to increase its speed from 20m/s to 40m/s.​

Chemistry
2 answers:
Grace [21]2 years ago
6 0

Answer:

\huge\boxed{\sf t = 50 \ seconds}

Explanation:

<h3><u>Given data:</u></h3>

Acceleration = a = 0.4 m/s²

Initial Speed = V_i = 20 m/s

Final Speed = V_f = 40 m/s

<h3><u>Required:</u></h3>

Time = t = ?

<h3><u>Formula:</u></h3>

\displaystyle a =\frac{V_f-V_i}{t}

<h3><u>Solution:</u></h3>

Rearranging formula for t

\displaystyle t =\frac{V_f-V_i}{a} \\\\t = \frac{40-20}{0.4} \\\\t = \frac{20}{0.4} \\\\\boxed{t = 50 \ seconds}\\\\\rule[225]{225}{2}

joja [24]2 years ago
4 0

Answer:

50 seconds

Explanation:

Equation used :

<u><em>v = u + at</em></u>

<u><em /></u>

Given :

⇒ a = 0.4 m/s²

⇒ u = 20 m/s

⇒ v = 40 m/s

Solving :

⇒ 40 = 20 + 0.4t

⇒ 0.4t = 20

⇒ t = 20/0.4

⇒ t = 50 seconds

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3 years ago
Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K sp = 1.4 × 10 –5). 1.5 × 10–2 M 1.4 × 10–
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Answer:

3.0x10⁻²M

Explanation:

Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:

Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻

When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:

ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:

1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

1.4x10⁻⁵ = [2X]² [X]

1.4x10⁻⁵ = 4X³

3.5x10⁻⁶ = X³

0.015 = X

As [Ag⁺] is 2X:

[Ag⁺] = 0.030 = 3.0x10⁻²M

The answer is:

<h3>3.0x10⁻²M</h3>
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3 years ago
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