Kinetic Energy = (1/2) mv^2.
m = 57.7 g = 57.7/1000 = 0.00577 kg.
v = 325 m/s.
E = 0.5 * 0.00577 * 325^ 2. Use your calculator.
E = 304.728125 J.
That's the kinetic energy.
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
Yes because all planets when they spin arund the sun they change season ,tho they change faster or slower then ours
deceleration or rėtardation i’m pretty sure (it won’t let me say the second word but it’s correct)
The area between the 10 and the 12.