True, the path of the ball, as observed from the train window, will be a horizontal straight line.
An object projected from a certain height has a parabolic path when observed from a fixed point.
However, if the reference point is moving at the same velocity as the object, the path of the object's motion appears to be a straight line.
When the ball is released from the window of the train, it will move at the same constant velocity as the train, and the path of the ball's motion observed from the train window will be a straight line.
Thus, we can conclude that the given statement is true. The path of the ball, as observed from the train window, will be a horizontal straight line.
Learn more about path of motion of objects here: brainly.com/question/82610
Time=50s
speed=25m/s
Distance = speed×time
=25×50
=1250m
DISTANCE TRAVELLED IS =1250m
Answer:8.2 Ω
Explanation:
I ASSUME you mean Ω (ohms) and not W(atts)
Re = 1/(1 / 27 + 1/56 + 1/15) = 8.2263329... ≈ 8.2 Ω
Answer:
1.) Covering the ceiling and walls with soft perforated boards
2.) Hanging curtains round the hall
3.) Having more opening in the wall.
Explanation:
This is due to what we called Reverberation due to poor acoustic properties.
Reverberation can be reduced by;
1.) Covering the ceiling and walls with soft perforated boards
2.) Hanging curtains round the hall
3.) Having more opening in the wall.
Answer:
a) h = 593.50 m
b) h₁₁ = 103 m
c) vf = 107.91 m/s
Explanation:
a)
We will use second equation of motion to find the height:

where,
h = height = ?
vi = initial speed = 0 m/s
t = time taken = 11 s
g = 9.81 /s²
Therefore,

<u>h = 593.50 m</u>
b)
For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

where,
vf = final velocity at tenth second = v₁₀ = ?
t = 10 s
vi = 0 m/s
Therefore,

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

where,
h = height covered during last second = h₁₁ = ?
vi = v₁₀ = 98.1 m/s
t = 1 s
Therefore,

<u>h₁₁ = 103 m</u>
c)
Now, we use first equation of motion for complete motion:

where,
vf = final velocity at tenth second = ?
t = 11 s
vi = 0 m/s
Therefore,

<u>vf = 107.91 m/s</u>