Answer:
a. 1222.13 J b. 2.36 m. His new position is above his original position
Explanation:
From work-kinetic energy principle, workdone = change in kinetic energy
So work the skateboarder does on himself W₁ = -116 J
work done by friction W₂ = -264 J
gravitational potential energy change W₃ = mgΔy
The kinetic energy change ΔK = 1/2m(v² - u²) where m = mass of skater = 52.9 kg, u = initial speed of skaterboarder = 2.04 m/s and v = final speed of skaterboarder = 6 m/s. ΔK = 1/2m(v² - u²) = ΔK = 1/2 × 52.9(6² - 2.4²) = 842.13 J
So, W₁ + W₂ + W₃ = ΔK
W₃ = ΔK - W₁ - W₂ = 842.13 J - (-116 J) - (-264J) = 842.13 J + 116 J + 264J = 1222.13 J
b. Since W₃ = mgΔy = 1222.13 J
Δy = W₃/mg = 1222.13/(52.9 × 9.8) = 1222.13/518.42 = 2.36 m
Since Δy > 0, his new position is above his original position.
A). 1,000 watts = 1 kilowatt
5,000 watts = 5 kilowatts
b). (5 kilowatts) x (2 hours) = 10 kilowatt-hours (kWh)
c). (10 kWh) x (15 cents/kWh) = $1.50
Typebars
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Sorry if this is not what you want me to do and I just threw a bunch of words at you.
<h2>Thus the force of friction is 235 N</h2>
Explanation:
When the bear was at the height of 14 m . Its potential energy = m g h
here m is the mass of bear , g is acceleration due to gravity and h is the height .
Thus P.E = 27 x 10 x 14 = 3780 J
The K.E of the bear just before hitting = m v²
= x 27 x ( 6.1 )² = 490 J
The force of friction f = P.E - K.E = 3290 J
Because the work done = Force x Distance
Thus frictional force = = 235 N
Answer:
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