(a) We must first look at the formulas of the velocities of each toy car. v1 =
-4.2 + 2.60t. v2 = 5.20. When the two cars have equal speed, then
v1 = v2
-4.2 + 2.60t = 5.20
2.60t = 9.40
t = 3.62 s
(b) Their speed would then be 5.20 m/s. The toy car does not change speed since it doest not have any acceleration.
(c) The two cars will pass each other when their positions are equal.
x1 = 13.5 - 4.2t + 0.5*2.60t^2
x2 = 8.5 + 5.20t
x1 = x2
13.5 - 4.2t + 1.30t^2 = 8.5 + 5.20t
1.30t^2 - 9.40t + 5.0 = 0
t = 6.65s or t = 0.58 s
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Kinetic energy<span> increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the </span>stopping distance<span> should increase by a factor of four, assuming that the driver is </span>can<span> apply the brakes with sufficient precision to almost lock the brakes.</span>
I always thought he was in it
Answer:
I = 0.287 MR²
Explanation:
given,
height of the object = 3.5 m
initial velocity = 0 m/s
final velocity = 7.3 m/s
moment of inertia = ?
Using total conservation of mechanical energy
change in potential energy will be equal to change in KE (rotational) and KE(transnational)
PE = KE(transnational) + KE (rotational)
v = r ω
I = 0.287 MR²