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1. ### A 68-kg fisherman in a 126-kg boat throws a package of mass m = 15 kg horizontally toward the right with a spe?

we are going to work this out by using momentum mass of fisherman = 126kg mass of package=15m speed of boat =0m/s speed of package= 4.6m/s momentum before = momentum after (126 x 0) ( 15 x 4.6) = (126v) ( 15 x 0) [ the package will eventually come to rest afterwards] 69= 126v 69/126= v v=0.55m/s...

2 Answers · Science & Mathematics · 02/05/2017

2. ### Hooke s Law and Spring Potential Energy?

...any point = total energy = kinetic energy + spring potential energy KE = 1/2*m*v^2 SPE = 1/2*k*x^2 1) KE = 1/4*Total Energy TE = 1/4...

1 Answers · Science & Mathematics · 22/04/2017

...25)m = -179 J The increase in spring potential energy is ΔSPE = ½kx² = ½ * 5.8N/m * (0.26m)² = 0.196...

2 Answers · Science & Mathematics · 09/11/2016

4. ### A slingshot consists of a light leather cup containing a stone.?

...3000 N/m When stretched 0.21 m, two bands would have SPE = 1/2K(0.21)² = 1500(0.21)² = 66.15 ≈ 66 J ANS...

1 Answers · Science & Mathematics · 03/08/2016

5. ### physics help?

...mass then: The energy stored in spring when displacement = 30 cm is: SPE = 1/2kx² = (0.5)(200)(0.30)² = 9 J The KE of...

1 Answers · Science & Mathematics · 25/07/2016

6. ### help with physics please?

...energy. GPE = mgh = 0.143kg * 9.80m/s² * (7.15+14.7)m = 30.6 J SPE = ½kx² = ½ * 877N/m * (0.429m)² = 80...

1 Answers · Science & Mathematics · 22/07/2016

7. ### (a) What is the momentum of a garbage truck that is 1.20×104 kg and is moving at 10.0 m/s ? (b) At what spe?

(a) p = mv = 1.2e5 kg·m/s (b) v = p / m = 1.2e5kg·m/s / 8kg = 1.5e4 m/s = 15 km/s

1 Answers · Science & Mathematics · 24/06/2016

8. ### Energy in an Oscillating Spring?

...left off units) The max spring pe occurs at the max displacement SPE Max = 1/2*k*A^2 = 1/2*m*w^2*A^2 All your...

1 Answers · Science & Mathematics · 11/06/2016

9. ### conservation of energy problem?

max SPE = 1/2kx² = (0.5)(600)(0.05)² = 0.75 J 0.75 = 1/2mV²...

2 Answers · Science & Mathematics · 15/04/2016

10. ### 6.4 calc problem?

SPE = 1/2kx² {where x = extended length in cm - 20} W1 = SPE...SPE(20) = (k/2)[(30-20)² - (20-20)²] = k/2(100) = 50k W2 = SPE(40) - SPE(30) = (k/2)[(40-20)² - (30-20)²] = k/2(400 - 100...

1 Answers · Science & Mathematics · 14/04/2016