I believe its the law of inertia
Vf = Vo + at
Vf = 20 m/s
Vo = 50 m/s
a = ?
t = 15
Therefore
20 = 50 + 15a
20 - 50 = 15a
-30 = 15a
a = -30 / 15
a = -2 m/s²
Answer:
Einstein extended the rules of Newton for high speeds. For applications of mechanics at low speeds, Newtonian ideas are almost equal to reality. That is the reason we use Newtonian mechanics in practice at low speeds.
Explanation:
<em>But on a conceptual level, Einstein did prove Newtonian ideas quite wrong in some cases, e.g. the relativity of simultaneity. But again, in calculations, Newtonian ideas give pretty close to correct answer in low-speed regimes. So, the numerical validity of Newtonian laws in those regimes is something that no one can ever prove completely wrong - because they have been proven correct experimentally to a good approximation.</em>
The discovery which Carnot made was that THE DIFFERENCE IN THE TEMPERATURES BETWEEN THE HOT AND THE COLD RESERVOIRS DETERMINE HOW WELL A HEAT ENGINE WOULD WORK.
Sadi Carnot was a French engineer, He proposed a theoretical thermodynamic cycle in 1824. In his cycle, Said hold that the efficiency of a heat engine depends on the temperature difference between its hot reservoir and cold reservoir.
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass 
kg, charge +e = 
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 
m/s. The proton comes momentarily to rest at a distance 
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 
m apart?
Explanation:
The given data is as follows.
Mass of proton = 
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
= 
where, 
U = 
Putting the given values into the above formula as follows.
U =
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .
