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Alika [10]
3 years ago
8

A 500 kg cart is rolling to the right at 1.3 m/s. a 60 kg man is standing on the right end of the cart. what is the speed of the

cart if tha man suddenly starts running to the left with a speed 10.0 m/s relative to the cart
Physics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

P1 = 1.3 (500 + 60) = 728 kg-m      total momentum to right at start

P2 = (v2 - 10) 60 + 500 v2

total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart

728 = 560 v2 - 600

v2 = 1328 / 560 = 2.37 m/s    new speed of cart

Check:

After:    p2 for cart = 500 * 2.37 = 1186

p1 for man = (2.37 - 10) * 60 = -458

P2 = p1 + p2 = 728       total momentum unchanged

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If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict
Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

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So,  

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n = 1

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Using ideal gas equation as:

PV=nRT

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\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT

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For Ar, given that:

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So,  

\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15

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