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:
The answer to the question is as follows
The acceleration due to gravity for low for orbit is 9.231 m/s²
Explanation:
The gravitational force is given as
Where = Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by and = 1 kg
rEarth = The distance between the two masses = 6.40 x 10⁶ m
therefore at a height of 400 km above the erth we have
r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m
and =
= 9.231 N
Therefore the acceleration due to gravity = /mass
9.231/1 or 9.231 m/s²
Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²
1) Maximum
2) Maximum
Explanation:
The force acting on a mass on a spring is given by Hooke's law; in magnitude:
where
F is the force
k is the spring constant
x is the displacement
Also we know from Newton's second law that we can write
where
m is the mass
a is the acceleration
So we can write the equation as
(1)
From this relationship, we see that the acceleration is directly proportional to the displacement.
On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by
(2)
where the first term is the elastic potential energy while the second term is the kinetic energy, and where
v is the velocity of the mass
From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.
Therefore this means that:
- When acceleration increases, velocity decreases
- When acceleration decreases, velocity increases
Therefore, the two answers here are:
- When the acceleration of a mass on a spring is zero, the velocity is at a maximum
When the velocity of a mass on a spring is zero, the acceleration is at a maximum