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3 years ago

A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?

Physics

1 answer:

Yanka [14]3 years ago

5 0

Answer:

$\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}$

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 380.2 \times {11}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 190.1 \times 121$

$\sf \implies KE = 190.1 \times 121$

$\sf \implies KE = 23002.1 \: J$

$\therefore$

Kinetic energy of the polar bear (KE) = 23002.1 J

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a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu

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2 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

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$h_A = 30 m$ (height at point A)

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$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

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Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

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