Answer:
The tension in the rope is 41.38 N.
Explanation:
Given that,
Mass of bucket of water = 14.0 kg
Diameter of cylinder = 0.260 m
Mass of cylinder = 12.1 kg
Distance = 10.7 m
Suppose we need to find that,
What is the tension in the rope while the bucket is falling
We need to calculate the acceleration
Using relation of torque


Where, I = moment of inertia
= angular acceleration

...(I)
Here, F = tension
The force is
...(II)
Where, F = tension
a = acceleration
From equation (I) and (II)


Put the value into the formula


We need to calculate the tension in the rope
Using equation (I)

Put the value into the formula


Hence, The tension in the rope is 41.38 N.
Answer:
990 J
Explanation:
Kinetic energy is:
KE = ½ mv²
Given m = 55 kg and v = 6 m/s:
KE = ½ (55 kg) (6 m/s)²
KE = 990 J
Answer:
<em>The total time is: t=451.22 sec</em>
<em>The average speed is: V=34.57 m/s</em>
Explanation:
<u>Average speed</u>
The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.
We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is
x=2*7,800 m=15,600 m
The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

The total time is:


The average speed is:


The maximum height to which the ball attain before falling back down is 1147.96 m
<h3>Data obtained from the question</h3>
The following data were obtained from the question:
- Initial velocity (u) = 150 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²
- Maximum height (h) =?
<h3>How to determine the maximum height </h3>
The maximum height reached by the ball can be obtained as illustrated below:
v² = u² – 2gh (since the ball is going against gravity)
0² = 150² – (2 × 9.8 × h)
0 = 22500 – 19.6h
Collect like terms
0 – 22500 = –19.6h
–22500 = –19.6h
Divide both side by –19.6
h = –22500 / –19.6
h = 1147.96 m
Thus, the maximum height reached by the ball is 1147.96 m
Learn more about motion under gravity:
brainly.com/question/22719691
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