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3 years ago

A 60.0 kg object is moving east at 8.00 m/s, and then slows down to 4.00 m/s. How much work was done?

Chemistry

1 answer:

Colt1911 [192]3 years ago

5 0

Answer:

-1440J

Explanation:

Hello,

This question requires us to calculate the work done on a object to move it from point A to point B

Data

Mass = 60kg

Initial velocity (V1) = 8.0m/s

Final velocity (V2) = 4.0m/s

Workdone on an object is equal to force applied on the object to move it through a particular distance.

Work done = force × distance

Force (F) = mass × acceleration

Distance = s

F = Ma

Work done = M× a × s

But a = velocity (v) / time (t)

Work done = mvs / t

But velocity = distance/ time

Work done = mv × v/

Work done = mv²

Work done = ½mv²

Workdone = ½M(V2² - V1²)

Workdone = ½ × 60 (4² - 8²)

Work done = 30 × (16 - 64)

Workdone = 30 × (-48)

Work done = -1440J

Work done = -1.44kJ

The workdone on the object is equal to -1.44kJ

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3 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$