Answer:
x₂ = 1.33 m
Explanation:
For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall
Στ = 0
W₁ x₁ - W₂ x₂ = 0
where W₁ is the weight of the woman, W₂ the weight of the table.
Let's find the distances.
Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.
x₁ = 2.5 -1.5 = 1 m
The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative
x₂ = 
let's calculate
x₂ = 
x₂ = 1.33 m
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
The freezing point of the water is 0 C , and it equals to 273 K
Then, To convert from Kelvins degrees to Celsius degrees we use the relation
Also,
We know that arc length (x(t)) is given with the following formula:
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
When we plug this back into the first equation we get:
We must solve this quadratic equation.
The final solution is:
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
