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Ivahew [28]
4 years ago
9

You've always wondered about the acceleration of the elevators in the 101-story-tall Empire State Building. One day, while visit

ing New York, you take your bathroom scale into the elevator and stand on it. The scale reads 150 lb as the door closes. The reading varies between 120 lb and 170 lb as the elevator travels 101 floors.
i)What is the maximum acceleration upward?
ii)What is the maximum magnitude of the acceleration downward
Physics
1 answer:
kondor19780726 [428]4 years ago
8 0

Answer:

(i) The maximum acceleration upward is 2.02 m/s².

(ii) The maximum acceleration downward is 1.39 m/s².

Explanation:

Let a be the maximum acceleration of the elevator.

The mass of the person at ground is 150 lb. We have to convert the mass into kg,

1 lb = 0.453592 kg

150 lb = 68 kg

170 lb = 77 kg

120 lb = 54 kg

(i) The person experience a force due to Earth's gravity in the downwards direction. The magnitude of this force is:

F₁ = mg = 68 x 9.8 = 666.4 N

The weight of the person decreases as the elevator is moving upwards. So, the force experienced by the person in this case due to gravity is:

F₂ = 54 x 9.8 = 529.2 N

Applying Newton's force equation;

(F₁ - F₂) = ma

(666.4 - 529.2) = 68 x a

a = 2.02 m/s²

(ii) The weight of the person increases as the elevator is moving downwards. So, the force experienced by the person in this case due to gravity is:

F₂ = 77 x 9.8 = 754.6 N

Applying Newton's force equation;

(F₁ - F₂) = ma

(666.4 - 754.6) = 68 x a

a = -1.30 m/s²

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In addition, it is the medium that will define, the propagation speed of the wave, according to its specific physical characteristics.

6 0
3 years ago
A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli
fredd [130]

Answer:

Part a)

A = 1.78 m

Part b)

f = 2 rev/s

Explanation:

Part A)

As we know that time period of the motion is given as

T = 2.68 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.68}

\omega = 2.34 rad/s

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

mg = m\omega^2 A

so we have

A = \frac{g}{\omega^2}

A = \frac{9.81}{2.34^2}

A = 1.78 m

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

mg = m\omega^2 A

g = \omega^2 (0.0623)

\omega = 12.5 rad/s

so now we have

2\pi f = 12.5

f = 2 rev/s

3 0
3 years ago
Un auto se mueve con velocidad constante de 20 m/s y la mantiene durante 20 s. Después se le aplica una aceleración constante y
pentagon [3]

Answer:

26.67 s

Explanation:

v = Velocidad final = 40 m/s

u = Velocidad inicial = 20 m/s

t_1 = Tiempo inicial = 20 s

t_2 = Tiempo empleado durante la aceleración

a = Aceleración

s = Desplazamiento del automóvil durante la aceleración = 200 m

De las ecuaciones lineales de movimiento tenemos

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{40^2-20^2}{2\times 200}\\\Rightarrow a=3\ \text{m/s}^2

v=u+at_2\\\Rightarrow t_2=\dfrac{v-u}{a}\\\Rightarrow t_2=\dfrac{40-20}{3}\\\Rightarrow t_2=6.67\ \text{s}

El tiempo necesario para acelerar es 6.67 s

El tiempo total necesario para toda la ruta es t=t_1+t_2=20+6.67=26.67\ \text{s}.

4 0
3 years ago
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley
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awnser :

Explanation:

8 0
3 years ago
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Find the components to write this vector in unit vector notation: 63.5 A ​please help
IrinaVladis [17]

Vector is perpendicular to x axis or i component.

Hence i component is 0

j component is 63.5

\\ \sf\longmapsto \overrightharpoon{A}=0\hat{i} +63.5\hat{j}

6 0
3 years ago
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