The 61.0 kg object<span> ... F = (300kg)(6.673×10−11 </span>N m<span>^2 </span>kg<span>^−2)(61kg)/(.225m)^2. F = 2.412e-5 </span>N<span> towards the 495 </span>kg<span> block. </span>b. [195kg] ===.45m ... (b<span>) You cannot achieve this </span>position<span>. For the </span>net force<span> to become zero, one or both of the </span>masses<span> must ...</span>
<span>there is no horizontal displacement if he went straight up
straight up means vertical, so his vertical displacment is 20 m</span>
Answer:
93 km/h
Explanation:
Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h
Let the first 5 hours journey distance = F
From the formula of speed,
Speed = distance/time
Substitute speed and time
72 = F/5
F = 72 × 5 = 360 km
The remaining distance will be:
639 - 360 = 279km
The remaining time will be:
8 - 5 = 3 hours
Speed = 279/3
Speed = 93 km/h
Therefore, the average speed for the remaining time of the journey is equal to 93 km/h
The answer for this question should be TRUE
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