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Vanyuwa [196]
3 years ago
13

A car travels 15 kilometers west in 10 minutes. After reaching the destination, the car travels back to the starting point, agai

n taking 5 minutes. What is the average velocity of the car?
A. 0 meters/seconds
B. 1.0 meters/seconds
C. 1.5 meters/seconds
D. 2.0 meters/seconds
Physics
1 answer:
jeka943 years ago
5 0

Speed = (distance traveled) / (time to travel the distance).
 
Strange as it may seem, 'velocity' is completely different. 

Velocity doesn't involve the total distance traveled at all. 
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there.  So the displacement in driving once around
any closed path is zero, because you end up where you started. 

Velocity =

           (displacement during some time)
divided by
            (time for the displacement)

AND the direction from the start-point to the end-point.


For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):

         Speed = (15km + 15km) / (10min + 5min)  =  (30/15) (km/min)

                                                                                 =  2 km/min.

        Velocity = (end location - start position) / (15 min)  =  Zero .

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7 0
3 years ago
You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be
Alex

Answer:

The amplitude  is  A =  90.2 \ m

Explanation:

From the question we are told that

    The frequency of when sound is approaching observer is   f = 392 Hz

     The frequency as the move away from observer  is  f_ a =  330 \ Hz

    The time between the pitch are t =  10 \ s

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

       T =  2 t

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

          T = \frac{2 \pi}{w}

Where  w is the angular velocity

=>   \frac{2 \pi}{w}  =  2 * 10

=>   w =  0.314 \ rad/sec

Now using Doppler Effect,

   The source of the sound is approaching the observer

The

          f = f_o (\frac{v}{v- wA} )

         392  = f_o (\frac{v}{v- wA} )

Where A is the amplitude

    So when the source is moving away from the observer

         f_a =  f_o (\frac{v}{v+ wA} )  

        330  =  f_o (\frac{v}{v+ wA} )  

Here  f_o is the fundamental frequency

Dividing the both equation  we have

           \frac{392}{330}  =  \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}

           1.1878  = \frac{v+wA}{v-wA}

         1.1878 v -  1.1878 wA = v+wA

        1.1878 v = 2.1878 wA

=>     A =  \frac{(0.1878 * (330))}{(2.1878)* (0.314)}

         A =  90.2 \ m

7 0
3 years ago
What happen when a star dies?
Vera_Pavlovna [14]
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Hope this helps!!
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3 years ago
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Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2
liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

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2 years ago
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