The "unstretched length" is shorter than the actual unloaded length because the load applied to the material caused the extension of the material thereby increasing its length.
<h3>Hooke's law</h3>
Hooke's law states that the force applied to an elastic material is directly proportional to the extension of the elastic material.
F = kx
where;
- F is the applied force
- k is the spring constant
- x is the extension
Thus, we can conclude that, the "unstretched length" is shorter than the actual unloaded length because the load applied to the material caused extension of the material thereby increasing the length of the material.
Learn more about Hooke's law here: brainly.com/question/2648431
By the law of universal gravitation, the gravitational force <em>F</em> between the satellite (mass <em>m</em>) and planet (mass <em>M</em>) is
<em>F</em> = <em>G</em> <em>M</em> <em>m</em> / <em>R </em>²
where
<em>• G</em> = 6.67 × 10⁻¹¹ m³/(kg•s²) is the universal gravitation constant
• <em>R</em> = 2500 km + 5000 km = 7500 km is the distance between the satellite and the center of the planet
Solve for <em>M</em> :
<em>M</em> = <em>F R</em> ² / (<em>G</em> <em>m</em>)
<em>M</em> = ((3 × 10⁴ N) (75 × 10⁵ m)²) / (<em>G</em> (6 × 10³ kg))
<em>M</em> ≈ 2.8 × 10¹⁴ kg
Answer:
a)The parachutist in the air for 12.63 seconds.
b)The parachutist falls from a height of 293 meter.
Explanation:
Vertical motion of parachutist:
Initial speed, u = 0m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 50 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 50
v = 31.32 m/s
Time taken for this
31.32 = 0 + 9.81 x t
t = 3.19 s
After 50m we have
Initial speed, u = 31.32m/s
Acceleration, a = -2 m/s²
Final speed , v = 3 m/s
We have equation of motion, v² = u² + 2as
Substituting
3² = 31.32² - 2 x 2 x s
s = 243 m
Time taken for this
3 = 31.32 - 2 x t
t = 9.44 s
a) Total time = 3.19 + 9.44 = 12.63 s
The parachutist in the air for 12.63 seconds.
b) Total height = 50 + 243 = 293 m
The parachutist falls from a height of 293 meter.