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Thepotemich [5.8K]
3 years ago
5

A tow truck uses a force of 100 N to pull a car 10 meters. How much work was done?

Chemistry
2 answers:
Rainbow [258]3 years ago
8 0
W = F * s
W = 100 * 10
W = 1000 J

In short, Your Answer would be 1000 Joules

Hope this helps!
olga55 [171]3 years ago
7 0

<u>Answer:</u> Work done by the tow truck is 1000 J

<u>Explanation:</u>

Work is defined as the amount of energy that is being transferred in order to move an object by an external force. It is expressed in joules.

The equation used to calculate amount of work done is:

W=F\times d

where,

W = Work done = ? J

F = Force applied = 100 N

d = Displacement of the car = 10 m

Putting values in above equation, we get:

W=100\times 10=1000J

Hence, work done by the tow truck is 1000 J

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The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?
n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of atmosphere = 5.00\times 10^{18}kg=5.00\times 10^{21}g

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = \frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol

Moles of carbon dioxide in air = \frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol

Hence, the moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

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3 years ago
A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?​
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0.6×30/298= 1×V2/273

V2=16.49L

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