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choli [55]
3 years ago
6

An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Gi

ve the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped. Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter ,- . If you think that both are zero, then you would enter 0,0 .
Physics
2 answers:
Oksanka [162]3 years ago
6 0

Answer:

Velocity: +ve, Acceleration: -ve

Explanation:

Here I've considered downward direction as positive direction.

OlgaM077 [116]3 years ago
5 0

Answer:

The answer is -,+ that is minus, plus

Explanation:

In the question, the elevator was described as moving downward, therefore its direction is negative. (-)  

From the question we could tell the elevator is decelerating, so the acceleration vector should be pointing upward, in contrast with the motion of the elevator.(+)

<h3>VELOCITY </h3>

Velocity is a vector quantity that indicates how fast an object is moving and in what direction, it has to do with and object’s displacement, time, and direction. The SI unit of velocity is meter per second (m/s). It should not be confused with speed which is a scalar quantity and measures on distance moved without stating what direction it moves.  

For instance, it would not be enough to say that the car has a velocity of 50 miles/hr. the direction in which the car moves must be included to fully describe the velocity of the car. The correct way would be the car has a velocity of 50 miles/hour East.

<h3>ACCELERATION </h3>

In physics, acceleration is defined as the rate of change of velocity. By altering an object’s speed or direction which changes its velocity hence its acceleration. Just like velocity, acceleration is a vector quantity. The SI unit of acceleration is meter per second squared (m/s^2)

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Answer:

p ≤ 8

Explanation:

Let's go over what we know that can help us build an inequality:

we know that these friends can spend no <em>more </em>than $325, meaning that they can either spend exactly $325, or they could spend less. This is our first clue, that one side of the inequality will be 325, and that the inequality sign will be <em>less than or equal to </em>( ≤ ) :

amount spent ≤ 325

Now, we want to figure out how many people can go (p)--but there's something we need to consider first--the parking.

Assuming that all friends take one car, we know we will have 11 dollars less to spend on admission (you can write this two ways).

amount spent on tickets + 11 ≤ 325

or, we can simplify it:

amount spent on tickets + 11 ≤ 325

                                          - 11     -11

amount spent on tickets ≤ 314

(I will be using the second way, you could always plug it back into the first way though)

now, we know that each "p" that we include will be multiplied by 39.25

         <em>Why?  Imagine we have one p (one person), they will cost 39.25, so we could write that as (p · 39.25) ,  ( 1· 39.25 ) -- we use × because that's how many </em><em>times</em><em> we're adding the number.  If we have two p, we could write it as (2p · 39.25)--there's an easier way to do this though</em>

<em />

So, we can write "p" to stand for all possible values of p possible,

p · 39.25 ≤ 314                  (we write this slightly differently):

39.25p ≤ 314          

Now, let's simplify!

   39.25p ≤ 314          

÷ 39.25     ÷ 39.25  (divide by 39.25 to isolate p)

          p   ≤   8

So, we know that the number of friends admitted must be less than or equal to 8, expressed as:

p ≤ 8

hope this helps! let me know if you need clarification on anything :)

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2 years ago
Read 2 more answers
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Newton's 2nd law of motion:   

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4 years ago
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mario62 [17]

Answer:

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Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

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For the electric field to be zero at the location

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\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

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\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

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4 years ago
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