If there are two equal and opposite forces on the SAME THING, then the thing doesn't accelerate. You're right about that. But the action and reaction forces act on two different things. The bullet and the rifle. The ball and the bat. The airplane and the air. etc. So BOTH can accelerate.
Hello!!
Here we have a simple matter of conservation of energy. ME=PE+KE.
At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.
Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!
Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
Where 
is the mass of the astronaut, 
is the mass of the satellite, 
is the speed of the satellite. We can calculate the speed 
of the astronaut:
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m
The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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