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stira [4]
2 years ago
7

A car travels up a hill at a constant speed of 44 km/h and returns down the hill at a constant speed of 74 km/h. Calculate the a

verage
Speed for the round trip.
Physics
1 answer:
Hunter-Best [27]2 years ago
5 0

Answer:

calculate it by yourself

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aev [14]
OD because Boyle’s law specifically states
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2 years ago
A child is sliding down a slide at the playground. is the mechanical energy conserved. why or why not.
aleksandrvk [35]
C: the mechanical energy isn't conserved. Some energy was lost to friction. 
6 0
2 years ago
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How much force is required to accelerate a 9.0-g object at 10000 g's?
Yuki888 [10]
Hey give us m = 9.0 g = 9.0 x 10-3 kg, and a = 10,000 "g's" = 98000 m/s/s so:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N
6 0
3 years ago
A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of
Anna007 [38]

Answer:

The ratio is  \frac{F_{T1}}{F_{T2}}  =  2

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       F_{T2} =  ma

 Considering the first car

 The force causing it s movement  is mathematically represented as

      F  = F_{T1} -F_{T2} = ma

=>   F_{T1} -ma  = ma

=>   F_{T1} =  2 ma

=> \frac{F_{T1}}{ma}  =  2

=> \frac{F_{T1}}{F_{T2}}  =  2

7 0
3 years ago
The input work done on a machine is 9.63 × 103 joules, and the output work is 3.0 × 103 joules. What is the percentage efficienc
notsponge [240]
Input work = 9.63×10³ J.
Output work = 3.0×10³ J

By definition,
Efficiency = (Output work)/(Input work)
                 = (3.0×10³)/(9.63×10³)
                 = 0.31 = 31%

Answer:  31%
5 0
3 years ago
Read 2 more answers
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