The strain energy stored in a linear spring is
SE = (1/2)*k*x²
where
k = the spring constant
x = the extension (or compression) of the spring
Given:
k = 470 N/m
x = 17.0 cm = 0.17 m
Therefore
SE = 0.5*(470 N/m)*(0.17 m)² = 6.7915 J
Answer: 6.8 J (nearest tenth)
Just as during formation, when the material contracts, the temperature and pressure increase. This newly generated heat temporarily counteracts the force of gravity, and the outer layers of the star are now pushed outward ( in not sure tho ) I hope this helps
Answer:
114.075 N
798.525 Nm
Explanation:
C = Drag coefficient = 0.5
ρ = Density of air = 1.2 kg/m³
A = Surface area = 9 m²
v = Velocity of wind = 6.5 m/s
r = Height of the tree = 7 m
Drag equation

Magnitude of the drag force of the wind on the canopy is 114.075 N
Toque is given by the product of force and radius

Torque exerted on the tree, measured about the point where the trunk meets the ground is 798.525 Nm
Answer:
Speed of lighter ball is 4 m/s.
Explanation:
Applying the principle of conservation of linear momentum,
momentum before collision = momentum after collision.

+
= 
- 

= 3 kg,
= 8 m/s,
= 2 kg,
= 0 m/s ( since it is at rest),
= 2 m/s,
= ?
(3 x 8) + (2 x 0) = (8 x 2) - (2 x
)
24 + 0 = 16 - 2
2
= 16 - 24
2
= -8
= 
= -4 m/s
This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.
THE answer the table is holding the book up would be a normal force