Now, you always beat him. Your grandfather is likely experiencing a slight decline in perceptual speed.
<u>Explanation:</u>
The speed of perception refers to the capacity to accurately (and completely) compare words letter, digits, objects, images, etc. When testing, these objects can be displayed simultaneously or one after the other. This type of test can be included in the proficiency test.
For example, we have also seen all the puzzles that ask the reader to notice the differences between the two pictures. The time it takes to recognize these differences is a measure of the speed of perception. Likewise, in getting rid of cards at the given situation, grandfather experiences a less decline in his perceptual speed.
- Initial velocity (u) = 0 m/s [the car was at rest]
- Distance (s) = 80 m
- Time (t) = 10 s
- Let the magnitude of acceleration be a.
- By using the equation of motion,
we get,
<u>A</u><u>nswer:</u>
<u>The </u><u>magnitude</u><u> </u><u>of </u><u>its </u><u>acceleration</u><u> </u><u>is </u><u>1</u><u>.</u><u>6</u><u> </u><u>m/</u><u>s^</u><u>2</u><u>.</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
it's because some versions have more steps and others have less
Answer:
Similarity: >>Time is independent variable and such is on the x-axis. ... >>Distance time graph tells you how much distance you have travelled, while velocity time graph tells you your acceleration. The difference between them is that the velocity-time graph reveals the speed of an object (and whether it is slowing down or speeding up), while the position-time graph describes the motion of an object over a period of time.
Explanation:
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g 
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g 
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
How energy is conserved
Em₀ = 
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)