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sergeinik [125]
3 years ago
12

A bobsled has a momentum of 4000 0 kg* m/s to the south. Friction on the track reduces its momentum to 500 kg* m/s to the south.

What impulse is applied to the bobsled by the friction?
A. 6500 kg* m/s north

OB. 1500 kg* m/s south O

C. 6500 kg* m/s south

O 1500 kg* m/s north
Chemistry
2 answers:
In-s [12.5K]3 years ago
7 0

Answer: B

Explanation: 4000- 1500

faltersainse [42]3 years ago
6 0
The correct answer is c
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Please review the attachment
astra-53 [7]

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Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

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Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

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Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

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