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galina1969 [7]
3 years ago
12

On a coordinate plane, vertex A for triangle ABC is located at (6,4). Triangle ABC is dilated by a scale factor of 0.5 with the

center of dilation at the origin. The resulting image is triangle A'B'C'. What are the coordinates of vertex A'? ​
Physics
1 answer:
BigorU [14]3 years ago
8 0
<h2>Answer:</h2>

A'(0.5\times 6,0.5 \times 4)=A'(3,2)

<h2>Explanation:</h2>

Dilation is when you stretch something by the same amount in two perpendicular directions. In other words, dilation changes size, not overall shape. If the factor of dilation is greater than one the shape is greater than the original, else if the scale factor is less than one the shape is smaller. In this problem, we must find the coordinates of A'. Since the center of dilation is the origins, we just need to multiply the coordinates of A by 0.5. So:

A'(0.5\times 6,0.5 \times 4)=A'(3,2)

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A 10.3 kg weather rocket generates a thrust of 240 N. The rocket, pointing upward, is clamped to the top of a vertical spring. T
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Answer:

(a) x = 0.25 m

(b) v = 1.46 m/s

(c) v = 2.4 m/s  

Explanation:

mass (m) = 10.3 kg

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spring constant (k) = 400 N/m

stretch distance from thrust (y) = 30 cm = 0.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) from mg = kx

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x = \frac{10.3 x 9.8}{400}

x = 0.25 m

   

(B) from the conservation of forces

  (Fy) + (0.5kx^{2}) =  (0.5ky^{2}) + mgh + (0.5mv^{2})

v = \sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2}) -  (0.5ky^{2}) - mgh }{0.5m}[/tex]}[/tex]

v =  \sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2}) -  (0.5 x 400 x 0.3^{2}) - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 1.46 m/s

                 

(C) if the rocket weren't attached to the spring, the conservation of energy equation becomes

 (Fy) + (0.5kx^{2}) = mgh + (0.5mv^{2})

v = \sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2})  - mgh }{0.5m}[/tex]}[/tex]

v =  \sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2})  - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 2.4 m/s                          

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