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Murrr4er [49]
3 years ago
10

Can someone help me ?

Physics
1 answer:
hammer [34]3 years ago
4 0

Answer:

1)    Time interval                 Blue Car                      Red Car

          0 - 2 s                Constant Velocity           Increasing Velocity

          2 - 3 s                Constant Velocity           Constant Velocity

          3 - 5 s                Constant Velocity           Increasing Velocity

          5 - 6 s                Constant Velocity           Decreasing Velocity

2) For Red and Blue car y₂  = 120       v = \frac{y_{2}-y_{1}}{t_{2}-t_{1}} = \frac{120-0}{6-0} = 20 m/s

     We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.

3)   At t = 2s, the cars are the same position, and are moving at the same rate

                    Position - same

                    Velocity - same

The position-time graph shares the same spot for two cars.

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In a water cycle Solid state of matter has the particles closest together.
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Read 2 more answers
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 7
Kisachek [45]

Explanation:

It is given that,

Total mass is 70 kg

The truck exerts a constant force of 20 N.

Then the net force is given by :

F = ma

a is acceleration of rider

a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2

Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :

v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s

Since, 1 m/s = 2.23 mph

4.28 m/s = 9.57 mph

So, the speed of the rider is 4.28 m/s or 9.57 mph.  

5 0
3 years ago
Find the velocity, acceleration, and speed of a particle with the given position function. r(t = t2i 6tj 4 ln t k
artcher [175]
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d r(t)/ dt = 2t i + 6 j + 4/t k

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6 0
3 years ago
Which requires more work, carrying a 420 N knapsack up a 200 m hill or carrying a 210 knapsack up a 400 m hill? Why?
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Answer:

b

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4 0
3 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
3 years ago
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