The time needed for the hammer to reach the surface of the Earth is 3.54 s.
<h3>
Time of motion of the hammer</h3>
The time of motion is calculated as follows;
t = √(2h/g)
where;
- h is height of fall
- g is acceleration due to gravity
t = √(2 x 10 / 1.6)
t = 3.54 s
Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.
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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:
True.
Explanation:
If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.
The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.
Hence, the given statement is true.
The power require to keep the car traveling is 6,666 W.
The power of the engine at the given efficiency is 3,999.6 W.
<h3>What is Instantaneous power?</h3>
This the product of force and velocity of the given object.
The power require to keep the car traveling is calculated as follows;
P = Fv

The power of the engine at the given efficiency is calculated as follows;

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