Answer:
0.66c
Explanation:
Use length contraction equation:
L = L₀ √(1 − (v²/c²))
where L is the contracted length,
L₀ is the length at 0 velocity,
v is the velocity,
and c is the speed of light.
900 = 1200 √(1 − (v²/c²))
3/4 = √(1 − (v²/c²))
9/16 = 1 − (v²/c²)
v²/c² = 7/16
v = ¼√7 c
v ≈ 0.66 c
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
Answer: when a object gets slowed down, it's force is either going into friction and drag, if it's on the ground, and weight+drag+friction, if it's in the air.
Explanation:
Answer:

Explanation:
From the conservation of mechanical energy




Solve to velocity v2




Answer:
The rate of change of distance is defined as speed.
Explanation:
The speed is defined as the rate of change of distance.
Speed = distance/ time
When we know the distance and the time, we get the value of speed. So, e know that who is moving fast or slow.
hen a graph is pltted beteen the distance and time, the slope of the graph gives the value of speed. So, by checking the slopes, hoseslope ismore, the speed is more and thusit is moving faster.
So, i agree with the statement.