Answer:
The distance of the object to the center of the earth increases.
Explanation:
The acceleration due to gravity on Earth is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance of the object from the Earth's centre
We notice that:
- g does not depend on the mass of the object
- g is inversely proportional to r
This means that if the distance of the object from the Earth's centre increases, g decreases. So, the correct option is
The distance of the object to the center of the earth increases.
Answer:
It does not hit the students face because the speed of the balloon slows down as energy is lost through thermal.
Explanation:
Answer:
I think no.2 the answer
Because socialization and social resources are both for me
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
