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Irina-Kira [14]

3 years ago

12

Write the answer to this question. Energy is measured in units called joules(J). Suppose a roller coaster car starts with 78,000

J of potential energy. In a few seconds, it converts two-thirds of this energy into kinetic energy. As it takes a curve, the car doubles its kinetic energy. How much mechanical energy does the car now have?

1 answer:

kolezko [41]3 years ago

7 0

Answer:78,000 J

Explanation:

Given

Initial Potential Energy P=78,000 J=78 kJ

Total Energy=Potential Energy+Kinetic Energy

initially kinetic energy is zero therefore

Total Energy=Potential Energy

If two-thirds of this energy converts to kinetic Energy

i.e. kinetic Energy $=\frac{2}{3}\times 78=2\times 26=52 kJ$

As it take curve its kinetic Energy doubles i.e. $2\times 52=104 kJ$

but total Energy is constant i.e. 78 kJ

therefore there must be a decrease in Potential energy of coaster

78=104+P.E.

P.E.=-26 kJ

Mechanical Energy of car is =sum of kinetic and Potential Energy

=78 kJ=78,000 J

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Explanation:

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3 years ago

. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the the perso

viktelen [127]

Answer:

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b) - 29.4 J

Explanation:

Given:

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Height above the floor, h = 1.5 m

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5 0

3 years ago

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

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$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

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$1024=-16t^2+64t+960$

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Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

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