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maria [59]
3 years ago
11

A group of big city tenants were upset about rising rent for their apartments, and decided to play a prank on the Building Manag

er. They went up to the top floor of the
building, which was 176.4 m above the sidewalk, and waited. They wanted to drop the balloon at just the right time, so they needed to make a quick calculation. How many
seconds before the Building Manager walks to the spot should they drop the balloon?
Physics
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:

  t = 6 s

Explanation:

This is a free fall exercise

          y = y₀ + v₀ t - ½ g t²

If the balloon is released its initial velocity is zero, when it reaches the floor its height is also zero, we substitute

         0 = y₀ + 0  - ½ g t2

         t = √(2yo / g)

let's calculate

         t = √ (2 176.4 / 9.8)

         t = 6 s

the balloon must be released 6 s before the person reaches the building

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Answer:s=0.68 m

Explanation:

Given

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deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

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3 years ago
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AleksAgata [21]

Answer:

a little

Explanation:

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When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second
4vir4ik [10]
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Divide both sides of the equation by 'acceleration', and you have

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8 0
2 years ago
Arocket launches at an angle of 33.6 degrees from the horizontal at a
babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, \theta=33.6

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A vector at an angle \theta with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle \theta with the horizontal, then the horizontal and vertical components are given as:

A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

u_y=u\sin \theta

Plug in the given values and solve for u_y. This gives,

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Answer:

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