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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 1000.0 kg truck accelerates from 20.0 m/s to 25.0 m/s over a distance of 300.0 m. What is the average net force on the truck?
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The average net force on the truck is 375 Newtons.

Explanation:

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Using these values in the above equation, we get acceleration = 0.375 m/s^{2}

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A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
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