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dedylja [7]
4 years ago
12

I need help on question 5

Physics
1 answer:
11Alexandr11 [23.1K]4 years ago
3 0
Weathering, erosion, impact, pressure  
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A woman lives on the eighth floor of an apartment building. She works in a high-rise office building 6.5 blocks away from her ap
san4es73 [151]

Answer:

b)   d = 997.9 m ,  c) D = (952.9 i ^ +45 k ^) m ,  D = 953.6 m and θ = 2.7º

Explanation:

In this exercise we must add the distance traveled, remembering that the displacement is a vector and the distance a scalar.

a) The displacement scheme is the woman walks in her building A to descend to the lower floor 4.0 m, we assume that this includes the vertical displacement, until reaching the street level, the displacement is vertical in this part.

Being on the street, she travels the 6.5 blocks to reach the building where she works, they indicate that each block is 146.6 m, this movement is horizontal.

Upon reaching building B, she goes up to his office on the 14th floor where she travels 5.5m on each floor, it is assumed that the distance to go up to the upper floor is included, this displacement is vertical

b and c) Let's find the distance traveled and the displacement

in building A

                          Zₐ = 8 * 4.0

                          Zₐ = -32.0 m k ^

the vector k ^ indicates that the displacement is vertical and the negative sign that it is descending

on the street

                        X_{ab} = 6.5 146.6

                        X_{ab} = 952.9 m i ^

the vector i ^ indicates that the displacement is the x-axis, we assume that the axis is in the direction of the displacement

in building B

                         Z_{b} = 14 * 5.5

                         Z_{b} = 77 m k ^

displacement in the vertical axis and in the positive direction

now we calculate the distance traveled,

               d = Zₐ + X_{ab} + Z_{b}

               d = -32 + 952.9 + 77

               d = 997.9 m

note that this value is a scalar

Let's calculate the displacement,

Z axis

           Z_{total} = Zₐ +  Z_{b}

           Z_{total} = -32 + 77

          Z_{total} = 45 m k ^

X axis

           X_{total} = X_{ab}

           X_{total} = 952.9 m i ^

we can give the result in two ways

a) D = X_total i ^ + Z_total k ^

    D = (952.9 i ^ +45 k ^) m

b) in module form and angles

Let's use the Pythagorean theorem

     D² = X_{total}^2 + Z_{total}^2

     D = √(952.9² + 45²)

     D = 953.6 m

We use trigonometry

     tan θ = Z / X

     θ = tan⁻¹ (Z / X)

     θ = tan⁻¹ (45 / 952.9)

     θ = 2.7º

this angle is measured from the positive side of the x axis towards the z axis

6 0
3 years ago
How much force would be needed to
Galina-37 [17]

There would be two forces acting on the box parallel to the floor, with a net force of

∑ <em>F</em> = <em>p</em> - <em>f</em> = <em>m a</em>

where <em>p</em> = magnitude of the push, <em>f</em> = mag. of friction, <em>m</em> = mass of the box, and <em>a</em> = acceleration. To find <em>p</em>, we first need <em>f</em> .

There are also only two forces acting on the box perpendicular to the floor, with net force

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

where <em>n</em> = mag. of normal force of the floor on the box and <em>w</em> = weight of the box. The net force is 0 because the box is only accelerating parallel to the floor.

<em />

<em>w</em> = <em>m g</em>, where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity, so we can solve for <em>n</em> :

<em>n</em> = <em>w</em> = <em>m g</em>

<em>n</em> = (22 kg) (9.8 m/s²)

<em>n</em> = 215.6 N

The kinetic friction is proportional to the normal force by a factor of the given coefficient of friction, <em>µ</em> = 0.17, such that

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.17 (215.6 N)

<em>f</em> = 36.652 N

Now solve for the required pushing force:

<em>p</em> - 36.652 N = (22 kg) (1.9 m/s²)

<em>p</em> ≈ 78 N

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3 years ago
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An ocean liner leaves New York City and travels 18.0o north of east for 155 km. How far east and how far north has it gone? In o
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