This an example of the law of Conservation of Energy.
The car has quite a bit of kinetic energy while it's rolling. If you want to stop it, you have to take that kinetic energy away from the car, AND you have to do something with that energy.
If it's an electric car or hybrid, you can turn the kinetic energy into electrical energy, put it back into the batteries, and use it again later.
If it's just an ordinary gas guzzler, there's no way to save the kinetic energy. You use the car's kinetic energy to scrape two rough surfaces together, that turns it into heat, and the air blows the heat away.
Next time you want the car to roll again, you have to make more, new, kinetic energy. So you take chemical energy out of more gas, and you use the motor to turn the chemical energy into kinetic energy.
It's all the law of Conservation of Energy ... in action.
3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:
(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v
where v is the velocity of the combined players. Solve for v :
450 kg•m/s - 320 kg•m/s = (155 kg) v
v = (130 kg•m/s) / (155 kg)
v ≈ 0.84 m/s
4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that
(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v
where v is the new velocity of the 4-kg ball. Solve for v :
30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v
v = (16.4 kg•m/s) / (4 kg)
v = 4.1 m/s
Answer:
A = 0.22 m
Explanation:
The spring with the block and the pebble forms an oscillatory system, this system is described by the expression
x = A cos (wt + φ)
w = √ (k / m).
The data they give us is the amplitude of the movement (A = 10 cm), the oscillation mass is equal to the block mass plus the mass of the pebble
m = m + M
m = 0.031 + 0.108
m = 0. 139 kg
To find the spring constant let's use Hooke's law
F = k X
The force is the weight of the pebble and the additional elongation is x = 4.9 cm, let's calculate
k = F / x
k = mg / x
k = 0.031 9.8 / 0.049
k = 6.2 N / m
Let's look for angular velocity
w = √ (6.2 / 0.139)
w = 6,670 rad / s
Let's write the oscillation equation with this data
x = 0.10 cos (6,670 t)
For the pebble to remain in contact with the block, the acceleration of the spring system plus block with pebble must be less than the acceleration of gravity in the descending oscillation
Let's look for system acceleration
a = d²x / dt²
dx / dt = - A w sin (wt + Ф)
d²x / dt² = - A w² cos (wt+Ф)
To find the maximum value cos (wt) = ±1
g = A w²
A = g / w²
A = 9.8 / 6.67²
A = 0.22 m
When the amplitude of the oscillation exceeds this value the pebble is delayed with respect to the block
Answer:
the mass deficit is directly proportional to bond reeling
Explanation:
The effect of the mass deficit in the nuclei is
δ = m_nucleons - m_nucleo
the delta quantity is the binding energy of the atomic nucleus,
This mass is related to energy by the Einstein relation
ΔE = c² Δm
therefore the mass deficit is directly proportional to bond reeling