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iren [92.7K]
3 years ago
15

A person walks 2 miles every day to work, leaving her front porch at 7:00 A.M. and arriving at work at 7:30 A.M. ON the way home

she walks the same route and ends at her front porch. What is her displacement?
Physics
1 answer:
Nataly [62]3 years ago
7 0

Answer:

The displacement is zero miles

Explanation:

The displacement of an object that moves from point A to point B is defined as

d =B-A

Where d is the displacement of the object. The displacement does not depend on the trajectory of the object. It only depends on the linear distance between the end point and the starting point.

In this case we know that the person walks from home to work and then walks from work to home. Therefore, the total displacement is the linear distance between the point where its journey begins and the point where the route ends.

The tour begins on the front porch of your house and ends on the front porch of your house (when you return from work). If we call A to the front porch of the house then the displacement is:

d = A-A = 0

The displacement is zero miles, since the person finishes the journey just where it started (front porch)

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Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
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Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

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The answer would be 420 m/s

Explanation:

Look in attachment ⬇

I Hope this Helps!!!

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Answer:

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What we can see from the formula is that, since the 2\pi does not change its value, the angular speed depends only on the period T.

In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.

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