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BigorU [14]
3 years ago
11

Discuss the environmental concerns regarding microfibre products

Engineering
1 answer:
zlopas [31]3 years ago
8 0

I don't know

Explanation:

cfffffffffggg

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A bicycle has tires that are 26 inches in diameter. The while in motion, the tires rotate through 125 complete revolutions. How
Stolb23 [73]

Answer:

850.8480103 feet

Explanation:

First you take the diameter and find the circumference, which is (2)(pi)(r) plug in your r which is 26/2= 13 so 2(13)(pi) and multiply taht by 125 after that take your answer and divide by 12which is 850.8480103

3 0
3 years ago
Sketch T-s and p-v diagrams for the Diesel cycle.
labwork [276]

Answer:

Diesel cycle:

    All diesel engine works on diesel cycle.It have four processes .These four processes are as follows

1-2.Reversible adiabatic compression

2-3.Heat addition at constant pressure

3-4.Reversible adiabatic expansion

4-1.Heat addition at constant volume

When air inters in the piston cylinder after that it compresses and gets heated due to compression after that heat addition take place at constant pressure after that power is produces when piston moves to bottom dead center.

From the diagram of P-v And T-s we can understand so easily.

3 0
3 years ago
If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

\zeta =\dfrac{C}{C_c}

Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks
Amiraneli [1.4K]

Answer:

  • 100N
  • 25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

4 0
3 years ago
Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.
nikitadnepr [17]

Answer:

a)  m_e= 3.05 Kg

b)  \rho=1072.3kg/m^3

c)  m_e= 3.05 Kg

Explanation:

From the question we are told that:

Beaker Mass m_b=1.20

Liquid Mass m_l=1.85

Balance D:

Mass m_d=3.10

Balance E:

Mass m_e=7.50

Volume v=4.15*10^{-3}m^3

a)

Generally the equation for Liquid's density is mathematically given by

m_e=m_b+m_l+(\rho*v)

\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}

\rho=1072.3kg/m^3

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

m_d=m- (\rho *v )

m=3.10+ (1072.30 *4.15*10^{-3}m^3 )

m=18.10kg

c)

Generally the equation for E's Reading at A pulled is mathematically given by

m_e=m_b+m_l

m_e = 1.20 + 1.85

m_e= 3.05 Kg

6 0
3 years ago
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