(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then
(140 kg) (15 m/s) = (140 kg + 200 kg) <em>v</em>
==> <em>v</em> ≈ 6.2 m/s
(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:
<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>
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Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Acceleration is the rate of change of velocity of an object. where a is acceleration, v is the final velocity of the object, u is the initial velocity of the object and t is the time that has elapsed. This equation can be rearranged to give: v = u + at.
-- In combination with 610 Hz, the beat frequency is 4 Hz.
So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.
In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.
-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.
So the unknown is (605+9) = 614 Hz.
