A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f
1 answer:
Answer:
Kf = 470 mJ
Explanation:
- According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
- Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
- Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:
- As we have already said, (1) is equal to the final kinetic energy of the puck:
- ⇒ Kf = 470 mJ (2)
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Answer:
(a) Elongation of the rod==5.61×10⁻⁹m
(b) Change in diameter=1.640×10⁻⁸m
Explanation:
Given data
Diameter d=78 in=1.9812 m
Cross Area is:
Applied Load P=17 KN=17×10³N
E=29 × 106 psi=1.99947961×10¹¹Pa
Stress and Strain in x direction
Stress in x direction
σ=P/A
σ=5517.25 Pa
Strain in x direction
ε=σ/E
ε=2.76×10⁻⁸
Part (a)
Elongation of the rod=Lε
=(0.2032)(2.76×10⁻⁸)
Elongation of the rod==5.61×10⁻⁹m
Part(b) Change in diameter
Strain in y direction
ε₁= -vε
ε₁= -(0.30)(2.76×10⁻⁸)
ε₁=-8.28×10⁻⁹
Change in diameter=d×ε₁
Change in diameter=(1.9812m)×(-8.28×10⁻⁹)
Change in diameter=1.640×10⁻⁸m
Answer:
Explanation:
Given that,
Emf, V = 22 mV
Number of turns in the coil us 519
Rate of change of current is 10 A/s.
We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.
Let's find the inductance first. So,
We have,
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is magnetic flux
So, the magnetic flux is .