Answer:
7.5 × 10^9 J.
Explanation:
So, we are given the following data or parameters or information for the proper solving of the question above;
=> ''actual force exerted by the tractor beam as a function of position is given by F(x) = ax + , where a = 6.1 x 10 N/mº and B = -4.1 x 10° N''
=> " Assume the supply spacecraft had an initial kinetic energy of KE = 2.7 x 10" J "
=> "that the tractor beam force is applied on the spacecraft over a distance of 7.5 x 10^11 m away.from its beginning position at = 0.0 m"
So, we can then solve it as;
KE2 – KE1 = F(x) dr.
KE2 - 2.7 x 10^11 J = - 3.5 × 10^6 × 7.5 x 10^4.
KE2 = 7.5 × 10^9 J.
Vfinal = Vinitial + at
Vinitial = 0
a = 9.8 m/s^2
t = 8s
Vfinal = 0 + 9.8 * 8
ANSWER: 78.4 m/s
I would say it would be the weight read.
Only because the Control is the type of liquid and the constants are Size of object and amount of liquid
Answer:
Explanation:
The given time is 1 / 4 of the time period
So Time period of oscillation.
= 4 x .4 =1.6 s
When the block reaches back its original position when it came in contact with the spring for the first time , the block and the spring will have maximum
velocity. After that spring starts unstretching , reducing its speed , so block loses contact as its velocity is not reduced .
So required velocity is the maximum velocity of the block while remaining in contact with the spring.
v ( max ) = w A = 1.32 m /s.
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
