Answer:
Both objects travel the same distance.
(c) is correct option
Explanation:
Given that,
Mass of first object = 4.0 kg
Speed of first object = 2.0 m/s
Mass of second object = 1.0 kg
Speed of second object = 4.0 m/s
We need to calculate the stopping distance
For first particle
Using equation of motion

Where, v = final velocity
u = initial velocity
s = distance
Put the value in the equation

....(I)
Using newton law

Now, put the value of a in equation (I)

Now, For second object
Using equation of motion

Put the value in the equation

....(I)
Using newton law


Now, put the value of a in equation (I)

Hence, Both objects travel the same distance.
25mph I hope this helps and sorry it took so long
Two hundred 432000 was broke down into aalll these parts
Transformation 3 should Tracy removed from the list. Chemical energy transforms into gravitational potential energy.
<h3> What is the law of conservation of energy?</h3>
According to the Law of Conservation of Energy, energy can neither be created nor destroyed, but it can be transferred from one form to another.
The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.
Transformations that occur as an engine runs to power a car is;
1. Chemical energy of fuel transforms into thermal energy.
2. Thermal energy transforms into electrical energy used to lighten the bulb in a car.
3. Thermal energy transforms into kinetic energy to help to run the car.
Chemical energy transforms into gravitational potential energy is the incorrect energy transformation.
Hence, transformation 3 should Tracy removed from the list.
To learn more about the law of conservation of energy, refer to brainly.com/question/2137260.
#SPJ1
Answer:
A. 
B. P ≈ 0
Explanation:
In order to calculate the magnetic field strength we have to use the magnetic field strength of a straight wire.
(eq. I)
B = magnetic field strength at distance d
I = current (A)
mi = represented by the greek letter μ, represents the permeability of the free space, which is: 4 × π 10^(-7) T m/A
d = distance from the wire
By replacing the values in eq I, we have the following:
(eq II)
The earth magnetic field in the surface variates from 25 to 65 microteslas. Thus:
P = Percentage from the wires/percentage of the earth
∵
∴
P ≈ 0