during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inc
1 answer:
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Answer:
a = 1.13×10^-8 m/s²
Explanation:
q = 15×10^-3 C
v = 2×10^6 m/s
B = 0.03 T
m = 8.00 x10^10 kg
a = 900/8 x10^10
a = 1.13×10^-8 m/s²
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