The bowling ball will require more force to roll because it is more massive.
That's efficiency. There's no law that it must be stated in percent.
As the plane falls the parabolic path remains directly below as the plane continues to fly over. This give more of an overview. When the package falls vertical acceleration happens as there is a vertical velocity as the package falls form high above. The downwards motion of gravity acts on the package if the approximated projectile motion ignoring air resistance.
Yes, peer pressure affects one's physical activity routine. It can do so both negatively and positively. For instance, if one is pressured to do drugs when around their peers, it would most likely lead to an addiction that lasts even when they are not with those people anymore. However, from a positive viewpoint, one's peers could also pressure them to do something productive, such as trying a new beneficial activity that they are afraid of (ex. trying out for a talent show.) This could lead to a disruption in routine as that individual would begin practicing for said talent show. Hence, peer pressure can be both negative and positive, but in both instances, it changes the routine of the individual effected.
Answer:
Somewhere between the two wires, but closer to the wire carrying λ₂
Explanation:
Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².
Electric Fied due to an electric charge is a vector and its direction is such that if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)
According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires are opposite
In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.
As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.
But for points closer to wire with λ₂ ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance to get equals E and then Ef = 0
