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Nataly_w [17]
3 years ago
9

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453

9kg. (a) If there is an uncertainty of 0.0001kg in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?
Physics
1 answer:
Drupady [299]3 years ago
8 0

Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

0.0001\times \frac{100}{0.4539}=0.022%

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

\frac{1}{0.022}\times {100}=4545.5\ kg

Mass in pounds would be

\frac{4545.5}{0.4539}=10014.32\ lb

Mass in pound-mass is 10014.32 lb

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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
2 years ago
You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica
Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A} , where:

l is the length

m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3

We divide the first equation by the second equation to get:

\frac{m}{R} = \frac{d A^2}{\rho}

A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8}   \ m^2

Using this Area, we find the diameter of the wire:

D = \sqrt{\frac{4A}{\pi}}

D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}

D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm

To find the length, we multiply the two equations stated initially:

mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}

l^2 = 8.534\\l =   2.92 \ m

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3 years ago
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Phase change from a liquid to a solid is what type of property change
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It is to freeze an object
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Answer:

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Explanation:

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