Answer: Approximately 22 cm
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Explanation:
The unstretched spring is 12.0 cm long. When adding a load of 5.0 N, it stretches to 15.0 cm. This is a displacement of 15.0 - 12.0 = 3.0 cm, which is the amount the spring is stretched.
Convert this displacement to meters (so that it fits with the meters unit buried in Newtons).
3.0 cm = (3.0)/100 = 0.03 m
Apply Hooke's Law to find the spring constant k
F = -kx
5.0 = -k*(0.03)
k = -(5.0)/(0.03)
k = -166.667 approximately
Now we must find the displacement x when F = 15 newtons
F = -kx
-kx = F
x = F/(-k)
x = -F/k
x = -15/(-166.667)
x = 0.089 approximately
x = 0.1
The displacement to one decimal place is about 0.1 meters, which converts to 100*0.1 = 10 cm
So the spring will be stretched to about 12cm+10cm = 22 cm
Answer:
Correct answer: A.) V = - 16.6 m/s down
Explanation:
Given:
V₀ = 3 m/s initial velocity
t = 2 seconds
g = 9.8 m/s²
V(t) = V(2) = ?
The movement described is a vertical upward shot
For velocity at any time is valid the next formula
V = V₀ - g · t
V = 3 - (9.8 · 2) = 3 - 19.6 = - 16.6 m/s down
Under condition that it has a enough drop height with respect to the ejection point.
God is with you!!!
Answer:
Some planets have seasons some don't bc of the distance from the sun some of them are too cold or too hot to have seasons
Momentum = mass x velocity
Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s
After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2
Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = <span>6.67 m/s
Total kinetic energy before
</span><span>= (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * ( -10 m/s) *2 ]
= 550 J
</span>
<span>Total kinetic energy after
</span>= (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s *2 ]
= 91.73 J
Total kinetic energy lost during collision
=<span>550 J - 91.73 J
= 458.27 J</span>
