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Paha777 [63]
3 years ago
13

A car travelling at 50m/s produces a force of 200N. What is the power of the engine in kilowatts(KW).

Chemistry
2 answers:
Zina [86]3 years ago
6 0

Answer:

10 kilowatts or 10,000 kw

Explanation:

Power = Force × Velocity

Given that:

Velocity of car is 50 m/s

The force is 200N

The power would be:

P = 200 × 50

P = 10,000 W

Notice that it only reaches watts because of the unit of the velocity is m/s. Standard unit to make Kw is km.

Since:

1 kw = 1000 w

10 kw = 10000 w

Credits to the person below me for correction.

zheka24 [161]3 years ago
5 0

Answer:

10 KW

Explanation:

The speed of the car is 50m/s.

The force of 200N is produced.

Power (W) = Velocity (m/s) × Force (N)

Power (W) = 50 × 200

Power (W) = 10000

1 Nm/s = 1 Watt

The power in Watts is 10000. The question asks for the power in kilowatts.

1 kilowatts = 1000 Watts

10000/1000 = 10 kilowatts

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umka2103 [35]

Answer:

94.44

Explanation:

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3 0
2 years ago
Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)
Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

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S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

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ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0
3 years ago
When the following equation is balanced using the smallest possible integers, what is the coefficent of oxygen gas?
elena55 [62]

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C7H16O + O2  →  7CO2 + H2O

You can make a table of the amount of atoms of C, H and O that you have in the reagents and products after you put a coefficient 7 in front of the CO2, in the following way,

R P

C 7 7

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Then you balance the hydrogens. It is better that you leave the last oxygens since there is an oxygen molecule alone, so when adding a coefficient to balance it, the quantities of the rest of the atoms in the equation would not be altered.

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The reaction and the table would be like this,

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To bring the coefficients of the reactants and the products to whole numbers multiply all the coefficients by 2. Then the reaction is like,

2 C7H16O + 21O2  →  14CO2 + 16H2O

As you can see when the given reaction is balanced using the smallest possible integers, the coefficent of oxygen gas is 21

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