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Neporo4naja [7]
2 years ago
11

A 1500 kg car

Physics
1 answer:
Lera25 [3.4K]2 years ago
5 0
<h2>We determine the final velocity, </h2><h2>v</h2><h2>f</h2><h2>, of the car. We do this by applying the kinematic equation,</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2 /><h2>where </h2><h2>v</h2><h2>i</h2><h2> is the initial velocity, a is the acceleration, and d is the distance covered across the movement. We can express the acceleration using the equation,</h2><h2 /><h2>a</h2><h2>=</h2><h2>F</h2><h2>m</h2><h2 /><h2>where F is the applied force and m is the mass of the car. We use the following values for the variables:</h2><h2 /><h2>v</h2><h2>i</h2><h2>=</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>F</h2><h2>=</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>m</h2><h2>=</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>d</h2><h2>=</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>We proceed with the solution.</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>Take the positive root.</h2><h2>v</h2><h2>f</h2><h2>=</h2><h2>√</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>=</h2><h2>√</h2><h2>(</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>)</h2><h2>2</h2><h2>+</h2><h2>2</h2><h2>(</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>)</h2><h2>(</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>)</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>≈</h2><h2>9.9</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>Help improve Study.com. Report an Error</h2>

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Answer:

The work done on the canister by the 5.0 N force during this time is

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Explanation:

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Answer:

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        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

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