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babunello [35]
3 years ago
9

Can be used to eliminate rubbing friction of wheel touching frame. 1.Traction 2.Thrust washer

Engineering
1 answer:
Vilka [71]3 years ago
5 0

Answer:

thrust washer

can be used to eliminate rubbing friction of wheel touching frame

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A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g
SSSSS [86.1K]

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

3 0
3 years ago
11. Ten bagels is what percentage of a dozen bagels?
IRINA_888 [86]

Answer:

the answer is d 83.3%

Explanation:

3 0
3 years ago
What unit of measurment would be used to measure current?
Alex_Xolod [135]

Answer:

The S. I unit of current is Amphere

5 0
3 years ago
A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is imp
Nikolay [14]

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

6 0
3 years ago
g A circular oil slick of uniform thickness is caused by a spill of one cubic meter of oil. The thickness of the oil slick is de
Anika [276]

Answer:

the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour

Explanation:

the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is

V = πR² * h

thus

h = V / (πR²)

Considering that the volume of the slick remains constant, the rate of change of radius will be

dh/dt = V d[1/(πR²)]/dt

dh/dt = (V/π) (-2)/R³ *dR/dt

therefore

dR/dt = (-dh/dt)* (R³/2) * (π/V)

where dR/dt = rate of increase of the radius , (-dh/dt)= rate of decrease of thickness

when the radius is R=8 m , dR/dt is

dR/dt = (-dh/dt)* (R³/2) * (π/V) = 0.1 cm/hour *(8m)³/2 * π/1m³ *(1m/100 cm)= 0.804 m/hour = 80.4 cm/hour

4 0
3 years ago
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