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AnnZ [28]
3 years ago
11

A softball is thrown from the origin of an X-Y coordinate system with an initial speed of 18 m/s at an angle of 35 degrees above

the horizontal. Part A) Find the X positions of the softball at the times t=.50s, 1.0s, 1.5s, and 2.0s. Part B) Find the Y positions of the softball at the times t=.50s, 1.0s, 1.5s, and 2.0s.
Physics
1 answer:
Goshia [24]3 years ago
7 0
Vo = 18 m/s
angle 35 degrees

1) Components of the initial velocity
 
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s

2) Equations of postion:

x = Vox*t
y = Voy*t - gt^2 / 2

3) Calculations

A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s

x = 14.74 * t

t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m

t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m

t = 1.5s => x = 22.11 m

t = 2s => x = 29.48 m

B)

y = Voy*t - gt^2 / 2

Voy = 10.32 m/s
g = 10 m/s (approximation)

y = 10.32*t - 5t^2

t = 0.5 s=> y = 3.91m

t = 1 s => y = 5.32m

t = 1.5 s => y = 4.23m

t = 2 s => y = 0.64 m



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Answer:

Correct answer: t = 2.86 seconds

Explanation:

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Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
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The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

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According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

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