In Newton's third law, the action and reaction forces D.)act on different objects
Explanation:
Newton's third law of motion states that:
<em>"When an object A exerts a force on object B (action force), then action B exerts an equal and opposite force (reaction force) on object A"</em>
It is important to note from the statement above that the action force and the reaction force always act on different objects. Let's take an example: a man pushing a box. We have:
- Action force: the force applied by the man on the box, forward
- Reaction force: the force applied by the box on the man, backward
As we can see from this example, the action force is applied on the box, while the reaction force is applied on the man: this means that the two forces do not act on the same object. This implies that whenever we draw the free-body diagram of the forces acting on an object, the action and reaction forces never appear in the same diagram, since they act on different objects.
Learn more about Newton's third law of motion:
brainly.com/question/11411375
#LearnwithBrainly
I believe the answer is b) slowly heating the surface
About 5 hours gooood luck
Answer:
Part 1) Time of travel equals 61 seconds
Part 2) Maximum speed equals 39.66 m/s.
Explanation:
The final speed of the train when it completes half of it's journey is given by third equation of kinematics as
where
'v' is the final speed
'u' is initial speed
'a' is acceleration of the body
's' is the distance covered
Applying the given values we get
Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as
Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance
Thus total time of journey equals
Part b)
the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 
Answer: 0.86 × 10^14
Explanation:
Given the following :
Radius of proton = 1.2 × 10-15 m
Radius of hydrogen atom = 5.3 × 10-11 m
Density of proton could be calculated thus:
Mass of proton = 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45
Density = mass / volume
Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)
= 0.2306 × 10^18
Density of hydrogen atom:
Mass of hydrogen atom= 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31
Density = mass / volume
Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)
= 0.2676 × 10^4
Ratio is thus:
Density of proton / density of hydrogen atom
0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14
