a simple lifting machine consisting of a rope which unwinds from a wheel on to a cylindrical drum or shaft joined to the wheel to provide mechanical advantage. reeeeeeeeeeeeeeeeeeeeeeeeee
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
So now we have an equation and unkown value.
for the thrown rock
for the dropped rock
solving both equation with the quadratic formula:
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
Answer:
a = 17.68 m/s²
Explanation:
given,
length of the string, L = 0.8 m
angle made with vertical, θ = 61°
time to complete 1 rev, t = 1.25 s
radial acceleration = ?
first we have to calculate the radius of the circle
R = L sin θ
R = 0.8 x sin 61°
R = 0.7 m
now, calculating at the angular velocity
ω = 5.026 rad/s
now, radial acceleration
a = r ω²
a = 0.7 x 5.026²
a = 17.68 m/s²
hence, the radial acceleration of the ball is equal to 17.68 rad/s²
Answer:
(D) 4
Explanation:
The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.
density = mass/volume
density = mass/(π(radius^2)(length))
So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.
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If you look at the maximum and minimum density, you find they are ...
{0.0611718, 0.0662668} g/(mm²·cm)
The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.
The error is about 4%.
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<em>Additional comment</em>
If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.
Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.
