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3 years ago

In 1985 in San Antonio, Texas, an entire hotel building was moved several

Physics

1 answer:

sleet_krkn [62]3 years ago

8 0

Answer:

The building was moved to a distance of, S = 510.2 m

Explanation:

Given data,

The mass of the building, m = 1.45 x 10⁶ kg

The work done to overcome the force of resistance, E = 1 x 10⁸ J

The force of resistance of the building, 2% of building weight

Fₓ = 2 x 1.45 x 10⁶ x 9.8 / 100

= 284200 N

The work done on the building,

E = Fₓ · S

∴ S = E / Fₓ

= 1.45 x 10⁶ / 284200 N

= 510.2 m

Hence, the building was moved to a distance of, S = 510.2 m

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Answer:

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3 years ago

What is the mass of an object that has a weight of 80.0 N?

Ira Lisetskai [31]

Explanation:

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3 years ago

Read 2 more answers

A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla

miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

$F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N$

The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.

$W = F_{net} *h\\\\W = 15.328 *10^{-3} * h$

W = K.E

$15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m$

Therefore, the ball traveled 0.827 m

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3 years ago