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

Primary mental skills needed for riding a motorcycle

Irina-Kira [14]

Answer:

You have to know the basic mechanics to help you correctly use the motorcycle, and you also have to have the right mindset - knowing you can ride it, and having determination. All of these factors are the primary mental skills you need for riding a motorcycle.

5 0

2 years ago

A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?

Illusion [34]

Answer:

0.0659 A

Explanation:

Given that :

$I_{0} = 6nA$ ( saturation current )

at 25°c = 300 k ( room temperature )

n = 2 for silicon diode

Determine the saturation current at 100 degrees = 373 k

Diode equation at room temperature = I = Io $\frac{V}{e^{0.025*n} }$

next we have to determine the value of V at 373 k

q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1

Given that I is constant

Io = $\frac{e^{0.025*2} }{31.08}$ = 0.0659 A

3 0

3 years ago

What is differences Between hard shoulder & soft shoulder in civil Engineerin?

r-ruslan [8.4K]

Answer:

The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'

Mark as brilliant........... 

5 0

3 years ago

An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60

belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F hi = 43.43 kJ/kg

WBT = 60 F wi = 0.00874 kJ/kg

u1 = 0.8445 m/kg

oulet condition,

DBT = 70 F RH = 100.1

h1 = 83.504kJ/kg

Wo = 0.222kJ/kg

check the attached file for complete solution

3 0

3 years ago

The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter

Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

( $p_t$ )₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ $[$ α₂/( $p_t$ )₂ $]^{1/2$ ------- let this be equation 2

where ( $p_t$ )₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ from equation for (σf)₂ in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ = 2(6( σ₀ )₁) $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

divide both sides by 2(σ₀)₁

$[$ α₁/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]$ = 36 $[$ 0.84/( $p_t$ )₂ $]$

1 / ( $p_t$ )₁ = 30.24 / ( $p_t$ )₂

( $p_t$ )₂ = 30.24( $p_t$ )₁

( $p_t$ )₂/( $p_t$ )₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0

3 years ago