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Ronch [10]
3 years ago
15

A small 20-kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy?

Physics
2 answers:
MariettaO [177]3 years ago
8 0
Ek=1/2mv^2 = (20)(1/2)(2x2) =40
Novosadov [1.4K]3 years ago
5 0

Answer: 40 Joules

Explanation: Kinetic energy is the energy possessed by a body by virtue of its motion.

Kinetic energy depends on the mass and speed of the object.

K.E=\frac{1}{2}\times mv^2

m= mass of canoe = 20 kg

v= speed of canoe = 2 m/s

K.E=\frac{1}{2}\times 20kg\times (2m/s)^2

K.E=40kgm^2s^-{2}=40Joules

(1kgm^2s^{-2}= 1Joule)

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Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

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(a) Xs=0.459m

(b) t=0.984 s

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(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

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