Explanation:
<h2><u>Steps </u><u>:</u></h2>
- <u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
- <u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
- <u>That's</u><u> </u><u>all </u>
<u>hope</u><u> it</u><u> </u><u>help</u>
<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>
Answer:
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from to is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area
As velocity is in and time is in so the unit of area is
Hence, the total distance is .
Answer:
The magnitude of the acceleration of the elevator is 0.422 m/s²
Explanation:
Lets explain how to solve the problem
Due to Newton's Law ∑ Forces in direction of motion is equal to mass
multiplied by the acceleration
We have here two forces 460 N in direction of motion and the weight
of the person in opposite direction of motion
The weight of the person is his mass multiplied by the acceleration of
gravity
→ W = mg , where m is the mass and g is the acceleration of gravity
→ m = 45 kg and g = 9.8 m/s²
Substitute these values in the rule above
→ W = 45 × 9.8 = 441 N
The scale reads 460 N
→ F = 460 N , W = 441 N , m = 45 kg
→ F - W = ma
→ 460 - 441 = 45 a
→ 19 = 45 a
Divide both sides by 45
→ a = 0.422 m/s²
<em>The magnitude of the acceleration of the elevator is 0.422 m/s²</em>