Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire,
=
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density, 
where,
=Avogadro's Number

Now,
the drift velocity, 

where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s
The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.
The correct answer is option D.
In the given graph, we can deduce the following;
- the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s
The average speed of the ant is calculated as;

The total distance from the graph is calculated as follows;
- first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm
- first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm
- second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm
- second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm
- third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm
- fourth downward distance from 12 cm to 9 cm = 3 cm
- final horizontal distance from 13 cm to 15 cm = 2cm
The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

The average velocity is calculated as the change in displacement per change in time.
The displacement is the shortest distance between the start and end positions.
- This shortest distance is the straight line connecting the start and end position. Call this line P
- From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
- Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm
Notice, you have a right triangle, now calculate the length of line P.
↓end
↓
↓ 6cm
↓
start -------------13 cm------------
Use Pythagoras theorem to solve for P.

The average velocity of the ant is calculated as;

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.
Learn more here: brainly.com/question/589950
Answer:
70m/s²
Explanation:
we will use the first equation of Dalton to find it
Answer:
7.74m/s
Explanation:
Mass = 35.9g = 0.0359kg
A = 39.5cm = 0.395m
K = 18.4N/m
At equilibrium position, there's total conservation of energy.
Total energy = kinetic energy + potential energy
Total Energy = K.E + P.E
½KA² = ½mv² + ½kx²
½KA² = ½(mv² + kx²)
KA² = mv² + kx²
Collect like terms
KA² - Kx² = mv²
K(A² - x²) = mv²
V² = k/m (A² - x²)
V = √(K/m (A² - x²) )
note x = ½A
V = √(k/m (A² - (½A)²)
V = √(k/m (A² - A²/4))
Resolve the fraction between A.
V = √(¾. K/m. A² )
V = √(¾ * (18.4/0.0359)*(0.395)²)
V = √(0.75 * 512.53 * 0.156)
V = √(59.966)
V = 7.74m/s