The resulatant displacement will be 8.60m
What is displacement?
Displacement is the shortest path covered by an individual.
We have given the distance 5.2km and 3.7Km.
The resultant will be given in the photo
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Answer:
F = 1480.77N
Explanation:
In order to calculate the required force to push the container with a constant velocity, you take into account the the sum of force on the container is equal to zero. Furthermore, you have for an incline the following sum of forces:
(1)
F: required force = ?
W: weight of the container = 1800N
N: normal force = weigth
α: angle of the incline = 28°
g: gravitational acceleration = 9.8m/s^2
μ: coefficient of friction = 0.4
You solve the equation (1) for F and replace the values of the other parameters:

The required force to push the container for the incline with a constant velocity is 1480.77N
1. One
2. Oohm
Hope this helps
Answer: There is no pic homie
Explanation:I am on the same question, so I can’t help :(
Answer:
6 bricks
Explanation:
For the bricks system to NOT fall over, then the center of mass of the system must lie within the touching area of the bottom brick.
Let the reference line be at the left end of the base brick, and the bricks are being stacking to the right direction.
- The 1st brick would have a center of mass at 20/2 = 10 cm or 10 + 0, this is < 20 cm so it stays
- 2 stacked bricks would have a center of mass at (10 + 14)/2 = 12 cm or 10 + 2, this is also < 20 cm so it stays
- 3 stacked bricks would have a center of mass at (10 + 14 + 18) / 3 = 14 cm or 10 + 2*2, this is also < 20 cm so it stays
- 4 stacked bricks would have a center of mass at (10 + 14 + 18 + 22)/4 = 16 cm or 10 + 2*3, this is also < 20 cm so it stays
- 5 stacked bricks would have a center of mass at (10 + 14 + 18 + 22 + 26)/5 = 18 cm or 10 + 2*4, this is also < 20 cm so it stays
- 6 stacked bricks would have a center of mass at (10 + 14 + 18 + 22 + 26 + 28)/5 = 20 cm or 10 + 2*5, this is also <= 20 cm so it stays before tipping over.
- The 7th brick would make everything fall down.