Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
Answer:
The magintude of the acceleration for both objects is 
Explanation:
Drawing a free body diagram on the two boxes we can analyze the system more easily.
we can take the acceleration going up as positive for reference purposes.
for mA let's suppose that is ascending so:
and for mB (descending):
because the two boxes has the same acceleration because they are attached together:
So the magintude of the acceleration for both objects is
Your answer is A
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Answer:
Simply,
<u>electrons</u> are "PARTICLES" orbiting the atoms, where, <u>current</u><u> </u>is the FLOW of some (free-to-move-around) electrons in a wire...
Answer:
Acceleration = 9 × 10^5 m/s^2 ( deceleration )
Explanation:
From the first equation of motion:
V = u + at
15000 = 30000 + 60a
a = ( 15000-30000)/60
a = 9 × 10^5 m/s^2
