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Marrrta [24]
3 years ago
15

Describe the motion of the object in Graph B of figure 11-2.

Physics
2 answers:
MatroZZZ [7]3 years ago
6 0

Explanation :

The attached graph is a distance-time graph for an object. The x-axis represents time while the y-axis represents the distance covered.

The motion can be uniform or non-uniform. When an object travels equal distances in equal intervals of time. The motion is uniform but when it does not cover equal distances in equal interval of time, the motion is non-uniform.

So, the motion in given graph B has different speed at every interval of time.

Hence, the motion of graph B is non-uniform.

kupik [55]3 years ago
4 0

Answer : The motion of the object is, non-uniform

Explanation :

Uniform motion : It is defined as the movement of an object along a straight line with constant speed.  It travels equal distances in equal time interval.

The average speed of an object is similar to the actual speed of an object.  The distance-time graph shows a straight line.

Non-uniform motion : It is defined as the movement of an object along a straight line with variable speed.  It travels unequal distances in equal time interval.

The average speed of an object is different to the actual speed of an object.  The distance-time graph shows a curved line.

The given graph B is a non-uniform motion graph.

Hence, the motion of the object is, non-uniform


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A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

dy = \sqrt{d^2 - dx^2}

dy = \sqrt{3.85^2 - 3^2}  = 2.41 km

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0
3 years ago
How long did the trip from camp wood to the pacific ocean and back again take?
maria [59]

The trip from Camp Wood to the Pacific Ocean and back again took 1.5 years to complete.<span>

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<span>The </span>Lewis<span> and Clark </span>Expedition<span> from May 1804 to September 1806, also known as the Corps of Discovery </span>Expedition, was the first American expedition<span> to cross what is now the western portion of the United States.</span>
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Why does the sound of a person’s voice sound different underwater than when they are speaking in air?
Tems11 [23]
Sound waves actually travel much faster in water than air, but words and the direction of the noise are distorted.
3 0
3 years ago
A stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s. What is the frequency perceived by
inn [45]

For a stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s, the frequency perceived  is mathematically given as'

F=81.721Hz

<h3>What is the frequency perceived by a firefighter racing toward the station at 11km/h?</h3>

Generally, the equation for the doppler effect  is mathematically given as

F'=\frac{vs+v}{vs}*f

Therefore

F=81(343+3.05556)/343

F=81.721Hz

In conclusion, the frequency is

F=81.721Hz

Read more about frequency

brainly.com/question/24623209  

4 0
2 years ago
A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4
SIZIF [17.4K]

Answer:

The  value is    v = 47 \  m/s

Explanation:

From the question we are told that

   The initial  speed of the roller coaster is u =  13 \  m/s

    The  length of the hill is  l   = 400 \  m

    The  acceleration of the  roller coaster is a=4.0 \ m/s^2

Generally the acceleration is mathematically represented as

      a =  \frac{ v - u}{ t_f -  t_i }

Here  t_i is the initial time which is equal to zero

         v_f is the final velocity which is mathematically represented as

          v_f  =  \frac{d}{ t_f}

So  

     a =  \frac{ \frac{d}{d_f}  - u }{ t_f - t_i}

     4 = \frac{\frac{400}{ t_f}  - 13}{t_f - 0}

      4 =  \frac{400 - 13t_f}{ t_f} *  \frac{1}{t_f}

     4t_f ^2  +13f  + 400 =

Solving this using quadratic formula we obtain

    t_f =  8.5 \ s

     t_f =  -11.8 \ s

Generally  time cannot be negative so

       t_f =  8.5 \ s

Generally the  final velocity is mathematically represented as

         v = \frac{400}{8.5}

         v = 47 \  m/s

       

5 0
3 years ago
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