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The actual mechanical advantage of a pulley is 2. The pulley has 3 supporting strings. What is the efficiency of the pulley?
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KGUS eNotes educator | CERTIFIED EDUCATOR
Hello.
The efficiency of a simple machine can be expressed as the ratio of the actual mechanical advantage divided by the theoretical mechanical advantage. You were given the actual mechanical advantage as 2, so you only need to determine the theoretical mechanical advantage. With a pulley system, this becomes a simple matter of determining how many strings support the pulley, which was nicely given as three. Therefore, the efficiency in this case is 2/3, or about 67%.
Answer:
Explanation:
Given
mass of spider
Largest amplitude can be obtained by Tapping after every 3 second
i.e. Time period of oscillation is
considering spider to execute Simple harmonic motion
time of oscillation is given by
substituting values
Answer: energy source, path, and load
Explanation:
lets assume
L = length of the object = 5 cm
h = height of the object = 3 cm
w = width of the object = 2 cm
V = Volume of the object
Since the dimensions of object suggest it to be a cuboid
Volume of the object is given as
V = L w h
inserting the values
V = 5 x 3 x 2
V = 30 cm³
m = mass of the object given = 10 g
ρ = density of the object = ?
density of the object is given as
density of object = mass of object / Volume of object
ρ = m/V
inserting the values
ρ = 10/30
ρ = 0.33 g/cm³
Answer:
(a) Approximately 968 Hz.
Explanation:
The observed frequency is less than 1003 Hz because of Doppler's Effect. When the source is moving away from an observer that doesn't move, the equation for the observed frequency would be:
,
where in the context of this problem,
- is the speed of sound in the air.
- is the speed at which the source moves away from the observer.
- is the frequency at the source.
Apply this equation to find :
.
Here's an alternative explanation.
The frequency of the siren at the source is . That corresponds to a period of .
In other words, at the source, a peak arrives about every .
The source is moving away from the observer at a speed of . In the between the first and the second peak, the source moved away from the observer. It would take an extra for the sound to cover that extra distance.
As a result, the period of the sound would appear to be to the observer.
That corresponds to an observed frequency of . (Same as the answer from the formula.)