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Vlad [161]
2 years ago
9

Write a conclusion to this lab in which you discuss when a person on a roller coaster ride would have sensations of weightlessne

ss and when they would have sensations of weightiness. In your discussion, talk about accelerations and forces. Then finish off your conclusion by using Newton's second law to explain why such accelerations and force conditions cause these sensations.
Physics
1 answer:
masya89 [10]2 years ago
3 0

Answer:

he lower part of the curve     N = M (g + v² / r)

upper part of the curve          N = m (v² /r -g)

Explanation:

In a roller coaster there is a long climb that allows the car to acquire gravitational potential energy, when this energy is converted into kinetic energy, there is a raven, in these curves we have two parts the lower part, where you have a feeling of great weight and another in the upper part where you have a feeling of weightlessness.

These sensations can be explained using Newton's second law, let's apply it to the lower part of the curve

         N-W = m a

acceleration is centripetal

        a = v² / r

we substitute

        N = mg + m v² / r

        N = M (g + v² / r)

In this part the apparent weight is increased by the speed of the body squared, it feels like a lot of fart.

In the upper part of the curve the force of gravity continues to act downwards, the normal that is the reaction of the surface also goes downwards, the centripetal acceleration pointing towards the center of the curve has a vertical downward direction

        -N -W = -m a

         N = ma -W

         N = m (v² / r -g)

In this case we see that the normal that gives the sensation decreases, which is why we feel a loss of weight, in the case of v2 / r = g, the request is total and the sensation of weightlessness.

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<span>Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:</span>

2 = k (2)^2

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k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

 

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

<span>d = 8 furlongs</span>

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A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
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<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

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Explanation

From the question we are told that

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    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

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Where dl is change of a unit length

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Where A is the area which is mathematically represented as

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  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

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