Answer:
a) w = 2.57 rad / s
, b) α = 3.3 rad / s²
Explanation:
a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground
Initial. Higher
Em₀ = U = m g h
Final. Touching the ground
= K = ½ I w²
How energy is conserved
Em₀ =
mg h = ½ I w2
The moment of specific object inertia
I = m L²
We replace
m g h = ½ (mL²) w²
w² = 2g h / L²
The height of the object is the length of the bar
h = L
w = √ 2g / L
w = √ (2 9.8 / 2.97)
w = 2.57 rad / s
b) the angular acceleration can be found from Newton's second rotational law
τ = I α
W L = I α
mg L = (m L²) α
α = g / L
α = 9.8 / 2.97
α = 3.3 rad / s²
Answer:
9 m
Explanation:
i did the test and got 100%
Potential energy increases as speed decreases. Kinetic increases when speed increases.
Given data:
- It is a graphical display where the data is grouped in to ranges
- A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
- It is an accurate representation of the distribution of numerical data.
<em>From Figure:</em>
Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).
<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>
Answer:
Fc = 89.67N
Explanation:
Since the rope is unstretchable, the total length will always be 34m.
From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:
L1+L2=34m
Replacing this value in the previous equation:
Solving for H:
We can now, calculate the angle between L1 and the 2m segment:
If we make a sum of forces in the midpoint of the rope we get:
where T is the tension on the rope and F is the exerted force of 87N.
Solving for T, we get the tension on the rope which is equal to the force exerted on the car:
