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Bingel [31]
3 years ago
7

A skateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angle of 60° above the horizontal. The end of the ra

mp is 1.5 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
Physics
1 answer:
erica [24]3 years ago
8 0

Answer:

y maximum   3.54 m, value X  2.35 m

Explanation:

We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero

                     

        Vyf² = Vyo² - 2g (Y-Yo)

Where Yo is the initial height of the ramp 1.5 m

        0 = Vyo² -2g (Y-Yo)

        Y-Yo = Voy² / 2g

       Y = Yo + Voy² / 2g

Let's calculate the velocity components using trigonometry

       Voy = vo without T

       Vox = Vo cost

       Voy = 7.3 sin 60

       Vox = 7.3 cos 60

       Voy = 6.32 m / s

       Vox = 3.65 m / s

Let's calculate the maximum height

         Y = 1.5 +6.32²/2 9.8

          Y = 3.54 m

This is the maximum height from the ground

b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point

                             

          Vfy = Voy - gt

          0 = Voy - gt

           t = Voy / g

           t = 6.32 / 9.8

           t = 0.645 s

     

Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time

          X = Vox t

          X = 3.65 0.645

          X= 2.35 m

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amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude <em>W</em> = <em>m g</em>

• the normal force, pointing perpendicular to the hill and away from the ground with mag. <em>N</em>

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector <em>F</em> pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. <em>F</em>.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = <em>W</em> (//) = <em>m</em> <em>a</em> (//)

∑ (⟂) = <em>W</em> (⟂) + <em>N</em> = <em>m </em><em>a</em> (⟂)

where, for instance, <em>W</em> (//) denotes the component of the sled's weight in the direction parallel to the hill, while <em>a</em> (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -<em>F</em> to the first equation.

If the hill makes an angle of <em>θ</em> with flat ground, then <em>W</em> makes the same angle with the hill so that

<em>W</em> (//) = -<em>m g </em>sin(<em>θ</em>)

<em>W</em> (⟂) = -<em>m g</em> cos(<em>θ</em>)

So we have

<em>-m g </em>sin(<em>θ</em>) = <em>m</em> <em>a</em> (//)   →   <em>a</em> (//) = -<em>g </em>sin(<em>θ</em>)

<em>-m g</em> cos(<em>θ</em>) + <em>N</em> = <em>m </em><em>a</em> (⟂)   →   <em>a</em> (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

<em>a</em> = <em>a</em> (//)

with magnitude

||<em>a</em>|| = <em>a</em> = <em>g </em>sin(<em>θ</em>).

6 0
3 years ago
When is a zero not significant?
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Answer:

Not between significant digits.

Explanation:

A zero not significant when it's not between significant digits.

8 0
3 years ago
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A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr
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Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds

The reaction time cannot be more than 1.69515 seconds

4 0
3 years ago
A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou
koban [17]

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

3 0
3 years ago
A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i
nexus9112 [7]

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135

4 0
3 years ago
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