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Diano4ka-milaya [45]
2 years ago
15

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically dr

ops that distance before the track curves forward. If the velocity at the bottom of the drop is 10.0 m/s and the mass of the cart and passengers is 3.5 x 104 kg, how much potential energy was converted into thermal energy (heat)
Physics
1 answer:
trapecia [35]2 years ago
5 0

The energies which is converted into thermal energy  is

2280.9 ×  10^4 J

According to the question when the ride will go upward then there will a potential energy acquired and when it comes down then kinetic energy starts to increase. since, the ride is starting from rest then kinetic energy will be maximum at lowest. So, now we will calculate P.E and K.E at various points.

Potential Energy at top when the vertical drop is 126.5 m is given as- P.E = m ×g× h

= 3.5 × 10^4 × 9.8 × 126.5

= 4338.95 × 10^4 J

Kinetic Energy at bottom when ride has velocity of 10 m/s then it is

given as -

K.E = 1/2 mv²

= 1/2 × 3.5×10^4 × 10²

= 175 × 10^4 J

Now the energy which converted to thermal energy is given as:

Thermal energy =  Potential energy - kinetic energy

= 2455.9 × 10^4 - 175 × 10^4

= 2280.9 × 10^4 J

Thus, the ride Potential energy was converted to thermal energy is 2280.9 × 10^4 J.  

Learn more about energies

here:brainly.com/question/12865479

#SPJ4

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A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

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(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

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E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

8 0
3 years ago
What is an instrument commonly used to measure wind speed?
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A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is
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I believe the correct answer from the choices listed above is the third option. <span>The force exerted by the book on the table is equal to the force exerted by the table which is 4.0 N. The book does not move so it must be that the forces are balanced. Hope this answers the question.</span>
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Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s
vichka [17]

Answer:

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In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

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Similarly in the second case probability = 2/12 = 1/6

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4 0
3 years ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

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Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

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0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
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