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gregori [183]
3 years ago
12

What does the phrase "1 meter per second per second" means? A. The velocity of an object can increase by 1 meter per second, for

every minute of time measured. B. The velocity of an object can increase by 1 meter per second, for every second of time measured. C. The velocity of an object can increase by 1 meter per second.
Physics
1 answer:
IRISSAK [1]3 years ago
6 0

Answer:

The answer is B.

Explanation:

We measure acceleration as a m/s2 because we are measuring the rate of change in velocity. Therefore, meters per second is the velocity of the object, and the second per second is the velocity added per second. For example, A vehicle is travelling at 75 m/s. Because it is on the road, it must stop at a light 150 meters ahead. To make it in time, the vehicle must apply a negative acceleration of at least 25 m/s2 to stop in time. The driver is in a hurry, so when the light turns green, the vehicle accelerates at a rate of 25 m/s2. That means that each second, the vehicle's velocity increases by 25 m/s. Make sense???

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Which of these forces determines the chemical properties of an atom? friction force nuclear force gravitational force electrical
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The answers is an electrical force.
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3 0
4 years ago
A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo
gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

                       H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

 Height at any time 'T' after the drop =

                          (initial height) +

                                              (initial velocity) x (T) +
                                                                 (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

     Upward  = the positive direction

                       Initial height = +150 m
                       Initial velocity = + 3 m/s

     Downward = the negative direction

                     Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

                 H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
0    =  (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

                   -4.9 T²  +  3T  + 150  =  0

Use the quadratic equation:

                         T  =  (-1/9.8) [  -3 plus or minus √(9 + 2940)  ]

                             =  (-1/9.8) [  -3  plus or minus  54.305  ]

                             =  (-1/9.8) [ 51.305  or  -57.305 ]

                          T  =  -5.235 seconds    or    5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

                    H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
H'(t)  =  v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set                                  v₀ + a T  =  0

                                      +3  -  9.8 T  =  0

Add 9.8 to each  side:   3               =  9.8 T

Divide each side by  9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh !  I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
                       (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) =  1.5 m/s .

Distance climbed during that time = (average speed) x (time)

                                                           =  (1.5 m/s) x (0.306 sec)

                                                           =  0.459 meter  (hardly any at all)

     But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at  150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely  5.847 seconds after its release.  

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work.  The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0
4 years ago
5. A construction worker on a high-rise building is on a platform suspended between two cables as illustrated below. The constru
Natalka [10]

Answer:

a) Tc = 750 [N]  ;b) See the explanation below.

Explanation:

To solve this problem, we first need a graphical explanation of this, as well as knowing the corresponding questions. Therefore, a search was carried out in google, in the attached image we will find a graphical description of the problem.

b)

The solution of this type of problem corresponds to the use of Newton's third law, applying static which tells us that the sum of the forces in a system in equilibrium without movement must be equal to zero.

a)

In this way we can find by means of a sum of forces on the y axis equal to zero:

- 850 - 450 + 550 + Tc = 0

Tc = 750 [N]

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