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cupoosta [38]
3 years ago
11

A person is standing on a scale placed on the floor of an elevator. At time t1, the elevator is at rest and the reading on the s

cale is 500N. At a later time t2, the person is still standing on the scale and the reading on the scale is 400N . Based on the scale readings, which of the following statements about the motion of the elevator could be true at time t2?
a. The elevator is moving upward at constant speed.
b. The elevator is moving upward and slowing down.
c. The elevator is moving downward at constant speed.
d. The elevator is moving downward and slowing down.
e. The elevator is at rest.
Physics
2 answers:
sergiy2304 [10]3 years ago
7 0

the following statements about the motion of the elevator could be true at time t2 is

c. The elevator is moving downward at constant speed.

Explanation:

  • Under given situation, the elevator is at rest and the reading on the scale is 500N.
  • After some time t2, the person is still standing on the scale and the reading on the scale is 400N .
  • It is because if you stand on a scale in an elevator which is accelerating upward, your body will feel heavier because the elevator's floor pressure which presses harder on your feet, and this is why the scale will show a higher reading than the time when the elevator is at rest.
  • Similarly on the other hand, when the elevator accelerates downward, your body will feel lighter. The force which is exerted by the scale is called as the apparent weight; which means it does not change with constant speed.
  • Applying Newton's second law, which concludes about this particular statement about force exerted on a body when at rest and when in motion.

Gennadij [26K]3 years ago
7 0

Answer:

id say a

Explanation:

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Answer:

100m \div  \frac{9 \: m}{1sec}  = 11.11 \: sec \:

I guess you can round it to 11 seconds.

Explanation:

Going with a speed 9m/s means you are going 9 meters in each second.

If you are going 9 meters in second how many seconds will it take to 100 meters?

Visually;

9 meters - - - 1 second

100 meters - - - ?seconds.

When you write like this 9 times ?seconds equal to 100 meters time 1 second. (you probably know this but just in case)

So to find ?second you multiply 100meters by 1 and divide it by 9 whixh will give you 11.1111 seconds whixh again I believe you can round it to 11.

(Kind of a) Proof;

If 9m * ?sec = 100 m * 1 sec

you send 9 meters to other side.

?sec = (100 m * 1 sec) ÷ 9m

Hope it was clear and it helps! Please let me know if you have any questions.

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The rate an object is moving relative to a reference point is its
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2 years ago
A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

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