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elena55 [62]
3 years ago
5

a canoe is floating down a river with a velocity of 35 meters per minute when it suddenly approaches a waterfall. if the waterfa

ll drop 30 meters, how far put will the canoe land?
Physics
1 answer:
vampirchik [111]3 years ago
3 0
From Newton's law v^2 = u^2 + 2as where a is the acceleration and s is the distance.  
But to go any further, we need to know how fast the vehicle is accelerating 
 From v = u +at
 We have a = u/t where the final velocity v = 0

 So in one minute acceleration = (35 / 60) / 60 = 0.0097 ms/2. The first
experession in bracket is the initial velocity, u, in metres per seconds. 
 Hence v^2 = (0.583)^2 + 2 (0.0097)(30)
 v^2 = 0.3398 + 0.5826 = 0.9224
 v = âš 0.9224 = 0.960m
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Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

q_{h} = R* T_{h} in(V_{2}/V_{1})

   =0.287 * 1200 ln(3)

   =0.287*1318.33

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The specific heat capacity:

q_{L}=q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

n_{TH}=1 - (T_{L}/T_{H})

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

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