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vichka [17]
3 years ago
6

A physics teacher stands at the edge of a cliff at Canyonlands National Park. The teacher wants to know how high above the base

of the cliff he is, so, making sure there are no hikers below drops a large rock over the edge of the cliff. 5.8 seconds after releasing the rock, the teacher hears the sound of the rock hitting the ground below. If sound can travel at a constant speed of 330 m/s, how high is the cliff?
Physics
1 answer:
S_A_V [24]3 years ago
5 0
 <span>Height = y 
T = 5.8 sec 
tf = time to fall 
ts = time for sound to travel distance y 

T = tf + ts = √[2y/g] + y/330 
or 
(T - y/330)² = 2y/g
 a quadratic in y. 
</span><span>y = 141.4 m
</span>hope this helps
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Answer:

The value is   a =  4.0 \ m/s^2

Explanation:

From the question we are told that  

  The velocity  is  a =  120 \  km/h =  [tex]\frac{120 *  1000}{3600} =  33.3 \  m/s[/tex]

  The time taken is t  =  8.28 \  s

    The  time taken for contact is  t_c  =  0.815 \  s

     The  velocity of the of the car after contact is v_c =  71.0 \  km/h = [tex]\frac{71 *1000}{3600} =  19.7 2 \  m/s[/tex]

From the equation of kinematics we have that  

       v =  u  + at

Here   u =  0 \ m/s  since the car is initially motionless

=>    33.3 =  0  + a *  8.28

=>    a =  4.0 \ m/s^2

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Answer:

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In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s

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let's calculate

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