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svp [43]
3 years ago
14

Desperado, a roller coaster built in Nevada, has a mass of 800 kg. It also has a vertical drop of 225 feet down the first hill.

The roller coaster is designed so that the speed of the cars at the end of this drop is 80 mi/h. Assume the cars are at rest at the start of the drop. How much work is done by friction on the car as it drops down the hill
Physics
1 answer:
weeeeeb [17]3 years ago
6 0

Answer:

the work done by friction on the car is 524,582 J.

Explanation:

Given;

mass of the roller coaster, m = 800 kg

distance moved by the coaster, d = 225 ft = 68.58 m

final velocity of the coaster, v = 80 mi/h = 35.76 m/s

The time taken for the coaster to drop down the hill is calculated as;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\ \times \ 68.58}{9.8} } \\\\t = 3.74 \ s

The work done by friction on the car is calculated as;

W =F\ \times \ d = \frac{mv}{t} \times \ d\\\\W = \frac{800 \ \times \ 35.76  }{3.74} \times \ 68.58\\\\W = 524,582 \ J

Therefore, the work done by friction on the car is 524,582 J.

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Musya8 [376]

Answer:

u₂ = 3.7 m/s

Explanation:

Here, we use the law of conservation of momentum, as follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\

where,

m₁ = mass of the car = 1250 kg

m₂ = mass of the truck = 2020 kg

u₁ = initial speed of the car before collision = 17.4 m/s

u₂ = initial speed of the tuck before collision = ?

v₁ = final speed of the car after collision = 6.7 m/s

v₂ = final speed of the truck after collision = 10.3 m/s

Therefore,

(1250\ kg)(17.4\ m/s)+(2020\ kg)(u_2)=(1250\ kg)(6.7\ m/s)+(2020\ kg)(10.3\ m/s)\\\\(2020\ kg)(u_2) = 8375\ N.s + 20806\ N.s - 21750\ N.s\\\\u_2=\frac{7431\ N.s}{2020\ kg}

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Answer:

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Explanation:

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Generally the equation for Satellite Speed is mathematically given by

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