Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ = 
L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
Answer:
16.6 °C
Explanation:
From the question given above, the following data were obtained:
Temperature at upper fixed point (Tᵤ) = 100 °C
Resistance at upper fixed point (Rᵤ) = 75 Ω
Temperature at lower fixed point (Tₗ) = 0 °C
Resistance at lower fixed point (Rₗ) = 63.00Ω
Resistance at room temperature (R) = 64.992 Ω
Room temperature (T) =?
T – Tₗ / Tᵤ – Tₗ = R – Rₗ / Rᵤ – Rₗ
T – 0 / 100 – 0 = 64.992 – 63 / 75 – 63
T / 100 = 1.992 / 12
Cross multiply
T × 12 = 100 × 1.992
T × 12 = 199.2
Divide both side by 12
T = 199.2 / 12
T = 16.6 °C
Thus, the room temperature is 16.6 °C
Answer:
<h3>1.01 s</h3>
Explanation:
Using the equation of motion S = ut+1/2gt² to solve the problem where;
u is the initial velocity of the chocolate = 0m/s
t is the time taken
g is the acceleration due to gravity = 9.81m/s²
S is the height of fall = 5.0m
Substituting the given parameter into the formula to get the time t we have;
5 = 0(t)+1/2(9.81)t²
5 = 4.905t²
t² = 5/4.905
t² = 1.019
t = √1.019
t = 1.009 secs
<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>
