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lyudmila [28]
3 years ago
14

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a consta

nt tangential speed of 1.81 m/s on its circular path. The rope holding the bucket unwinds without slipping on the barrel of the crank. Find the linear speed with which the bucket moves down the well.
Physics
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

0.453 m/s

Explanation:

Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is \pi d_h where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is \pi\times 0.4\approx 1.256 m

The rate of rotation will then be 1.81/1.256=1.441 rev/s

Similarly, circumference of inner part will be \pi d_i where subscript i represent inner. Substituting 0.1 for inner diameter then

\pi\times 0.1\approx 0.3142 m

The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s

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3 years ago
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A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant
nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

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3 years ago
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Describe an experiment to show that pressure increases with the decrease in the area of surface
cluponka [151]
Answer:
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7 0
2 years ago
Q1. Tractors are often used on sloping fields, so stability is important in their design. On the diagram, the center of the X ma
Vesna [10]

Answer:

a) The tractor has not toppled over because the the center of mass is acting in between the wheels

b) The features which affect its stability are;

1) The location of the center of gravity of the tractor close to the ground

2) The increased spacing between the left and the right wheels of the tractor

Q2. a) The center of mass wheel act in the region to the right away the wheel span (space in between the wheel) which creates a clockwise effect to, topple the buses over

b) It raises the bus's center of mass higher

c) To raise the center of mass to its highest practical level

Q3. At the bottom middle location within the tractor

Explanation:

a) The tractor has not toppled over because the vertical line from the center of gravity to the ground is still in between the wheel base such that since the tractor will topple over by turning clockwise about the right tire due to its weight, the weight of the tractor is currently acting in the anticlockwise direction to the right end tire of the tractor keeping the tractor's tires in good contact with the ground

b) The features which affect its stability are;

1) The location of the center of gravity of the tractor close to the ground

2) The increased spacing between the left and the right wheels of the tractor

Q2. a) When either of the buses are tilted further, the mass of the buses is acting on the tires and in their tilted condition, the mass is acting mainly on the right tire, whereby as the vertical line from the center of mass to the ground is in between the wheel bases the mass of the buses still serve s the restoring force to keep the bus on ground by providing anticlockwise moment where clockwise lotion is required for the bus to topple over

If the buses are tilted further, the vertical line from the center of mass will cross to the other side of the right tire such that the mass now provides clockwise moment turning the buses clockwise and since there are no opposing anticlockwise moment to balance that of the now clockwise moment of the buses it will continue to turn clockwise with the result that they eventually  topple over

b) When the upper deck is full of passengers, there will be an appreciable proportion of the total mass of the bus in the upper deck such that the level of the center of mass will be raised up higher

c) Sand bags are only put upstairs because by allowing for the disproportionate distribution of load such that the majority of the load of the bus is concentrated upstairs, the center of mass will be at its maximum height allowing the bus to be tested for stability in the worst case scenario

Therefore, the sands are only put upstairs to raise the center of mass

Q3. Being that the tractor is being used on a rough ground, the safest position will be at the lowest possible location at the middle of both the front and rear wheels and the left and right tires.

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3 years ago
Steam is leaving a pressure cooker whose operating pressure is 20 psia. It is observed that the amount of liquid in the cooker h
Debora [2.8K]

Answer:

a) 34.05ft/s

b).1156.2BTU/lbm

c) 2.04BTU/s

Explanation:

Amount of liquid that has evaporated, m = ◇Vliq/ Vf

We replace the values to make conversion

m = (0.6gal/ 0.01683ft^3/lbm) × (0.13368ft^3/1gal)

m = 4.755lb

The mass flow rate of exit steam is given by:

m' = m/◇t

We replace values to make conversion

m' =( 4.766lb/45min) = 0.1059lb/min × 1min/60s

m' = 0.001765lb/s

The exit velocity V = m'/pA = m'Vg/A

We replace values to make conversion

V =[ (0.001765lbm/s)(20.093ft^3/lbm) /(0.15 in^2)]× (144in^2/1ft^2)

V = 34.05ft/s

b) The total and flow energies per unit mass is given by:

Eflow= Pv = h - u

We replace the values to make conversion

Eflow = 1156.2 - 1081.8

Eflow = 74.4BTU/lbm

Therefore theta= h + ke + pe

Theta approximately =h = 1156.2BTU/lbs

c) The rate at which energy is leaving the cooler by steam is given by:

Emass = m'theta

Emass = (0.001765)×(1156.2)

Emass = 2.04BTU/s

6 0
3 years ago
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