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MariettaO [177]
3 years ago
7

At a highway speeds, a particular automotive is capable of of an acceleration of about 1.6m/s^2. At this rate how long does it t

ake to accelerate from 80km/h to 110km/h.
I know the formula would be v=v0+at. I just don't understand the converting. I would love an explanation on how to do this. I been stuck for a while.
Physics
1 answer:
8_murik_8 [283]3 years ago
3 0
Fine.  You just need a conversion at some point between seconds and hours.
Let's do it this way:

First, let's review one very useful definition of acceleration:

                   Acceleration = (change in speed) / (time for the change)

The question is asking for a time, so let's massage this equation to get
time all by itself.
Multiply each side by (time for the change) :

                   (Acceleration) x (time for the change) = (change in speed)

Divide each side by (acceleration) :

                   Time for the change = (change in speed) / (acceleration).

There it is.  The problem gives you the change in speed and the acceleration.

                    Change in speed = (110 - 80) = 30 km/hr

                    Acceleration = 1.6 m/s²

Here's where the only difficulty shows up.  One is in hours and the
other one is in seconds.
The way you change the units of a quantity is to multiply it by
a fraction whose value is ' 1 '.  That way you don't change the value
of the original quantity.

How does that help ?  You just pick the fraction very carefully !  Remember
that the value of a fraction is ' 1 ' if the numerator and denominator are both
the same value.  Here are some fractions that are equal to ' 1 ' :

    12 inches/1 foot     60 seconds/1 minute     1 meter/100 centimeters

See how that works ?  Multiply one of these by a quantity, and it changes
the units of the quantity, but not its value.

Let's change  1.6 m/s²  to m/hr² .

1 hour = 3,600 seconds  so the fraction is  (3600 sec/hr) .

The unit conversion is:  (1.6 m/s²) x (3600 sec/hr) x (3600 sec/hr)

I multiplied by the fraction twice because we need sec² on top to cancel
the s² on the bottom.

Multiply this out and you get  20,736,000 m/hr². 
Don't let the big number scare you.

Now go back to our equation for the time:

                    Time for the change = (change in speed) / (acceleration)

Change in speed = 30 km/hr

Acceleration was 1.6 m/sec²  but we found that
that's the same as 20,736,000  m/hr² .
And also, since the change in speed is in km,
let's change this one to  20,736 km/hr² .

                     Time for the change = (30 km/hr) / (20,736 km/hr²)

                     30 / 20,736 = 0.0001447 hour

                                      =  5.2 seconds.

I realize that this has turned into a nightmare.  I probably lost you, and
the answer is likely wrong because the numbers got so ugly.  I apologize. 
The reason for all the ugly numbers is that I apparently picked the wrong
number to change the units.  It probably would have been a lot neater to
convert the change in speed to  m/sec  instead of changing the acceleration
to  m/hr² .  But the principle is the same.  Make a fraction out of the unit you
have and the new unit you want, wit the same value on top and bottom, and
use it to convert the units of one number to match the units of the others.
========================================

Change in speed = (30 km/hr) x (1000 m/km) x (1 hr / 3600 sec) = 8.333 m/sec

Time = (change in speed) / (acceleration) = (8.333 m/s) / (1.6 m/s²) = <em><u>5.2 seconds</u></em>

Yes, this way was certainly better and prettier.
But at least I got the same answer going the long tough way!


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A car is moving with a constant velocity of 25 m/s. Which of the following is true?
S_A_V [24]

b) the net force on the car is zero.

Explanation:

Let's analyze each option one by one:

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a = 0

and so Newton's second law becomes

\sum F = ma = 0

where \sum F is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.

b) the net force on the car is zero.  --> TRUE, for what we said at point A)

c) the inertia is changing.  --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.

d) the forces of friction are proportional to the acceleration of the car.  --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is

F_f = \mu mg

where \mu is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.

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Learn more about net force and Newton's second law:

brainly.com/question/3820012

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6 0
3 years ago
The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant
uysha [10]

Answer:

The amount of charge on the plates is now equal to half its original  value

Explanation:

From the question we are told that  

  The charge on the plate is  Q

   The  initial  separation is d_1

    The new separation is d_2 = 2 d_1

Generally the capacitance of the capacitor is mathematically represented as

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Generally the charge of the parallel plate capacitor is mathematically represented as

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=>  Q =\frac{\epsilon *  A  *  V   }{ d}

Here  \epsilon ,  A and  V are constant , so looking at the question we see that Q varies inversely with d

  So  

       Q = \frac{K}{d}

Here K  is  a constant

so

     Qd =  K

=>  Q_1 d_1 =  Q_2 * d_2

=>   Q_1 d_1 =  Q_2 * [2 d_1]

=> Q_2 = \frac{Q_1}{2}

So we see from the mathematically relation that the charge of the parallel plate capacitor will reduce by half of it original  value

     

   

3 0
3 years ago
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