Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How
1 answer:
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
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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