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solmaris [256]
3 years ago
15

What is the total mechanical energy of a 200 kg roller coaster moving with a velocity of 16 m/s at a height of 18 m above the gr

ound?
Physics
1 answer:
Travka [436]3 years ago
8 0

Mechanical Energy = PE + KE

PE: mgh = 200 x 9.8 x 18 = 35280

PE: 35280 Joules

KE: 1/2mv^2 = 1/2 x 200 x 16^2 = 25600

KE: 25600 Joules

ME: 35280 + 25600

ME: 60,880J

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if the accepted thickness of aluminum foil is 1.5x10^5 is your answer percise, accurate or both. explain your answer
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Answer:

The answer is 150,000

Explanation:

I got my answer by solving 10^5 which is 100,000 then you just multiply 1.5 to get 150,000

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A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases he
juin [17]

The skater's final angular speed is equal to 12 rad/s.

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7 0
2 years ago
Explanation's of E=MC²
natita [175]

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Read 2 more answers
An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the
poizon [28]

Answer:

The value is A  =  0.014 \  m

Explanation:

From the question we are told that

    The mass of the object is  m  =  2.0 \  kg

    The unstressed length of the string is  l  =  0.08 \  m

    The length of the spring when it is  at equilibrium is  l_e = 5.9 \  cm  =  0.059 \  m

      The initial speed (maximum speed)of the spring when given a downward blow v  =  0.30 \  m/s

Generally the maximum speed  of the spring  is mathematically represented as

           u =  A *  w

Here A is maximum height above the floor (i.e the maximum amplitude)

            and w is the angular frequency which is mathematically represented as

       w = \sqrt{\frac{k}{m} }

So

        u =  A *   \sqrt{\frac{k}{m} }

=>      A  =  u *   \sqrt{\frac{m}{k} }

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

           b  =  l -l_e

=>         b  = 0.08 - 0.05 9

=>         b  = 0.021 \  m

Generally at equilibrium position the net force acting on the spring is  

            k *  b  -  mg  =  0

=>         k *  0.021   -   2 * 9.8  =  0

=>        k =  933 \  N/m

So

            A  =  0.30  *   \sqrt{\frac{2}{933} }

=>          A  =  0.014 \  m

8 0
3 years ago
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None I mean you have to take a drivers test but you will have to take the knowldege test twice

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