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3 years ago

In a rack and pinion steering system, what component protects other

Engineering

2 answers:

madreJ [45]3 years ago

8 0

Answer:

Explanation:

d- rubber bellows protects the system from any form of contamination

Alexxandr [17]3 years ago

7 0

Answer:

C or Rack

Explanation:

A) Gear housing

B) Pinion shaft gear

C) Rack <----

D) Rubber bellows

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Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic

Tom [10]

Answer:

Explanation:

The pictures below shows the whole explanation for the problem

4 0

4 years ago

A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?

Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is $L=100\ cm$

Strain experienced is $\epsilon=82\%=0.82$

Strain is the ratio of the change in length to the original length

$\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm$

Therefore, new length is given by (Considering the load is tensile in nature)

$\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm$

Thus, option (b) is correct.

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3 years ago

Given frequency, what is the formula for the period of a wave?

Veronika [31]

Answer:

f = c / λ = wave speed c (m/s) / wavelength λ (m). The formula for time is: T (period) = 1 / f (frequency). λ = c / f = wave speed c (m/s) / frequency f (Hz). The unit hertz (Hz) was once called cps = cycles per second.

Explanation:

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3 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

3 years ago

Read 2 more answers

Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th

Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0

4 years ago