Answer:
I think the answer is A) Power
Explanation:
The centripetal force is the gravitational force of attraction between the Earth and the Moon.
Also, the gravitational force of attraction between the Earth and the Sun is the centripetal force that causes the Earth to stay in orbit around the Sun.
Also, the gravitational force of attraction between a TV satellite, a weather satellite, or a GPS satellite and the Earth is the centripetal force that causes each satellite to stay in orbit around the Earth.
__Protons and neutrons have similar mass.
__Electrons are smaller than a proton or a neutron.
Explanation:
The three particles involved in this problem are:
- Proton: it is positively charge, it is found in the nucleus of the atom, and its mass is

- Neutron: it has no electric charge, it is also found in the nucleus of the atom, and its mass is approximately equal to that of the proton (just slightly larger)
- Electron: it has negative electric charge, it orbit around the nucleus of the atom, and its mass is much smaller than that of the proton:

We can now analyze each of the given statement:
__Protons and neutrons have similar mass. --> TRUE
__Protons and electrons have similar mass. --> FALSE, the electron is much lighter
__Neutrons and electrons have similar mass. --> FALSE, the neutron is much heavier
__Protons are smaller than a neutron or an electron. --> FALSE, protons are similar to the neutrons
__Neutrons are smaller than a proton or an electron. --> FALSE, neutrons are similar to the protons
__Electrons are smaller than a proton or a neutron. --> TRUE
Learn more about atoms:
brainly.com/question/2757829
#LearnwithBrainly
Answer:
Part a)

Part b)

Part C)

Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as



now we will have




Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have


Part C)
By momentum conservation we will have



Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.