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lora16 [44]
3 years ago
15

A mass m = 4.7 kg hangs on the end of a massless rope L = 2.06 m long. The pendulum is held horizontal and released from rest. 1

)How fast is the mass moving at the bottom of its path?
Physics
1 answer:
AysviL [449]3 years ago
6 0

Answer:

PE = KE

mgh = ½mv²

v = √(2gh)

v = √(2(9.81)2.07)

v = 6.37 m/s

Explanation:

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Answer:

T_{f}  = 17º C

Explanation:

This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)  

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     Q₁ = 2,664 10⁴ J

Now let's see what this liquid water temperature is when this heat is released  

      Q = M c_{e} ΔT = M c_{e} (T₀₁ -T_{f1})  

      Q₁ = Q  

     T_{f1} = T₀₁ - Q / M ce  

     T_{f1} = 26.0 - 2,664 10⁴ / (0.860 4186)  

     T_{f1} = 26.0 - 7.40  

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The initial temperature of the liquid water is T₀₁= 18.6  

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         T_{f} = (M To1 - m To2) / (m + M)  

         T_{f} = (0.860 18.6 - 0.080 0) / (0.080 + 0.860)  

T_{f} = 17º C

 

gg

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