Save image for better view. A cart running on frictioniess air tracks is propell

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A cart running on frictioniess air tracks is propelled by a stream of water expelled by a gas-powered pressure washer stationed on the cart. There is a 1.00-m3 water tank on the cart to provide the water for the pressure washer. The mass of the cart, including the operator riding it, the pressure washer with its fuel, and the empty water tank, is 405 kg. The water can be directed, by switching a valve, either backward or forward. In both directions, the pressure washer ejects 205 L of water per min with a muzzle velocity of 26.7 m/s. If the cart starts from rest, after what time should the valve be switched from backward (forward thrust) to forward (backward thrust) for the cart to end up at rest? Tries 0/25 What is the mass of the cart at that time? Tries 0/25 What is the velocity of the cart at that time? Tries 0/25 What is the thrust of this "rocket" before the valve is switched (enter first) and after the valve is switched Tries 0/25 What is the acceleration of the cart immediately before the valve is switched? Tries 0/25

Explanation / Answer

1) v = Ve*ln[(Mv+Mp)/Mv]
Let
Mv = 405 kg
Mp = 1000kg
Ve = 26.7 m/s
? = 205/60 = 3.41kg/sec

The accelerations will have 2 different times, t1 & t2. t1+t2 = Mp/? = 341 sec or,
2) t2 = 341 - t1

For the + acceleration, dv1 = Ve*ln[(Mv+Mp)/(Mv+Mp - ?*t1) = Ve*ln[1405/(1405 - 3.41*t1)]


For the - acceleration, dv2 = Ve*ln[(Mv+Mp - ?*t1)/(Mv+Mp - ?(t2+t1)]; Using (2), this becomes
......................................... Ve*ln[(1405 - ?*t1)/(1405 - 1000)]
Letting dv1 = dv2 ?

1405*405 = 1405

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