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3 years ago

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During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the de

celeration of a descending elevator, the normal force on the feet of a passenger is _____ her weight.

1 answer:

gizmo_the_mogwai [7]3 years ago

7 0

Answer: Smaller than ; larger than

Explanation:

When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of

N= mg +ma, (g is gravity and a is acceleration)

here ma is negative so the N= mg-ma

Hence, it feels smaller than its original weight.

When the elevator is moving downward , then the force acting will be positive in nature

N= mg+ma,

here ma will be positive so it feels larger the original weight of passenger.

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A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air

never [62]

Answer:

Approximately $2.1\; \rm km$ , assuming that $g = -9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $t$ denote the time required for the package to reach the ground. Let $h(\text{initial})$ and $h(\text{final})$ denote the initial and final height of this package.

$\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})$ .

For this package:

Initial height: $h(\text{initial}) = 2500\; \rm m$ .
Final height: $h(\text{final}) = 0\; \rm m$ (the package would be on the ground.)

Solve for $t$ , the time required for the package to reach the ground after being released.

$\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}$ .

$\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}$ .

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at $95\; \rm m \cdot s^{-1}$ .) From calculations above, the package would travel forward at that speed for about $22.588\; \rm s$ . That corresponds to approximately: $95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km$ .

Hence, the package would land approximately $2.1\; \rm km$ in front of where the plane released the package.

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4 years ago

What does the bohr model look like?

sergeinik [125]

Pic of Bohr Model attached

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4 years ago

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Which list shows the development of atomic models in chronological order?

ikadub [295]

I think it would be plum pudding, solid ball, solar system.

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3 years ago

A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from

kaheart [24]

Answer:

zero

Explanation:

In this system, only conservative forces act. Therefore, the mechanical energy, that is, the sum of the kinetic energy and the potential energy, remains constant. When the mass is at its maximum displacement from equilibrium, its potential energy is maximum, therefore, its kinetic energy is minimal, that is to say, that its instantaneous velocity at that point is zero.

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4 years ago

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

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3 years ago