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Maksim231197 [3]
3 years ago
14

an object travels 16m in 4sec and then another 16m in 2sec. what is the average speed of the object?​

Physics
2 answers:
Veronika [31]3 years ago
7 0

Answer:

Explanation:

Average Speed = Total Distance / Total Time

5.3 m/s

Equation

Total distance = d1 + d2

Total Time = t1 + t2

Givens

Total Distance = 16 + 16

Total Distance = 32

Total Time = t2 + t1

Total Time = 2 sec + 4sec

Total Time = 6 sec

Solution

Average Speed = Total Distance / Total Time

Average Speed = 32 / 6

Average Speed = 5.33 m/s

Annette [7]3 years ago
3 0

Answer:5.4

Explanation:

distance=rate*time

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The coefficient of static friction between a book and the level surface it slides on is 0.65. If the mass of the book is 0.2 kg,
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A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If th
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Answer:

The car is 72.75 miles away from its starting position.

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First, remember the relation:

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D = √( (a - c)^2 + (b - d)^2)

For this problem, we can assume:

The North is equivalent to the y-axis, and the East is equivalent to the x-axis.

We also assume that the initial position of the car is (0mi, 0mi)

Now the car moves to the East at a speed of 52mi/h for one hour, then the new position of the car is:

(0mi, 0mi) + (52mi/h*1h, 0mi) = (52mi, 0mi)

Now the car travels 30 mins (or 0.5 hours) to the northeast at a speed of 52mi/h.

We can assume that it moves at an exact angle of 45° from East to North, then the components of the speed can be written as:

Sx = speed in the x-axis = 52mi/h*cos(45°) = 36.77 mi/h

Sy = speed in the y-axis = 52mi/h*sin(45°) =  36.77 mi/h

Then the new position of the car is:

(52mi, 0mi) + (36.77 mi/h*0.5h, 36.77 mi/h*0.5h) = (70.385 mi, 18.385 mi)

Now we know the final position of the car.

The distance between the final position (70.385 mi, 18.385 mi) and the initial position (0mi, 0mi) is:

D = √( (70.385 mi - 0mi)^2 + (18.385 mi - 0mi)^2) = 72.75 mi

The car is 72.75 miles away from its starting position.

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