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3 years ago

The pressure of a car can be adjusted using what

Engineering

1 answer:

IgorLugansk [536]3 years ago

3 0

Answer:

tyre pressure monitoring system (TPMS)

Explanation:

It very important to inflate your tyres to the specified pressure. So cultivate a habit to check and refill them once a month. Note that you can't tell if a tyre is underinflated just by looking at it. If it actually looks underinflated in a way. And you can't really rely on a tyre pressure monitoring system (TPMS) only to keep track. Most systems only warn you when the pressure is 25 percent below the manufacturer's recommended tyre pressure.

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Which symbol should be used for the given scenario?

Firdavs [7]

Answer:

speed square

Explanation:

4 0

3 years ago

a horizontal jet of water (at 100c) that is 6 cm in diameter and has a velocity of 20 m/s is deflected by the vane as shown. if

Reptile [31]

The net resultant direct force and angle on the vane is created when the water jet exits the vane at position 2 with 92% of its initial velocity.

<h3>What is mean by velocity?</h3>

The speed at which a body or object is moving determines its direction of motion. A scalar quantity, speed is primarily. As a matter of fact, velocity is a vector quantity.

The rate at which distance changes is what it is. It measures the displacement's rate of change. A body's velocity is defined as its speed in a particular direction.

Velocity is a measure of how quickly a distance changes in relation to time. Having both magnitude and direction, velocity is a vector quantity.

The rate of change in a body's displacement with respect to time is referred to as velocity. In the SI, m/s is its unit.

Given,

External angle of Curved Vane = 158°

mean velocity at 1 = 12 m/s

Volumetric flow rate = $55 \mathrm{~m}^3 / \mathrm{h}=\frac{55}{3600} \mathrm{~m}^3 / \mathrm{s}$.$

mean velocity at $2=12 \times 0.92=11.04 \mathrm{~m} / \mathrm{s}$

i) Force exerted in x - friction A C 1 = Volume

$F_{S_x} &=\rho A C_1\left[C_2 \cos \theta-C_1\right] \\$

$&=1000 \times \frac{55}{3600}\left[\left(11.04 \cos 158^{\circ}\right)-12\right]$

i $\rangle F_{\text {sc }}=\supseteq A c_1\left[C_2 \sin \theta\right] \\$

$&=1000 \times \frac{55}{3600} \times \text { TI. 04 } \sin (1589 \\$

$&F_{\text {syn }}=63.18 \mathrm{~N} \\$

$&\text { Angle } \Rightarrow \frac{F_{s y}}{F_{3 x}}-\tan \theta \\$

$&\tan \theta=\frac{63.18}{339.18}, \theta=160-10-5.3 \\$

$&\theta=\tan ^{-}\left(\frac{-63 \cdot 18}{339728}\right) \\$

$&\theta=-10.540^{\circ} \\$

The complete question is:

A horizontal jet of water strikes a curved vane as shown in Figure C.1. The external angle of the curved vane is 158°.The mean velocity and volumetric flow rate of the water jet at position 1 are 12 m/s and 55 m³/h respectively. Due to friction, the water jet leaves the vane at position 2 with 92 % its original velocity.

(i) Direct force exerted by the water jet on the vane in the x - direction.

(ii) Direct force exerted by the water jet on the vane in the y - direction.

(ii) Net resultant direct force and angle on the vane.

To learn more about velocity, refer to:

brainly.com/question/24681896

#SPJ4

4 0

2 years ago

The type of model that is a complete and unambiguous mathematical representation of a precisely enclosed and filled volume is a(

gayaneshka [121]

A digital solid is a 3D model consisting of vertices, edges, faces, and partially filled or entirely filled interior. It is a complete and unambiguous representation of the object in a precisely enclosed and filled volume in digital space

5 0

3 years ago

An angle is observed repeatedly using the same equipment and procedures producing the data below:32 ∘ 37' 00", 32 ∘ 37' 10", 32

ki77a [65]

Answer:

a) x_average = 34.51077°, b) σ = 2.69°, c) σ = 2.5107°

Explanation:

a) The most likely value of a measure is

x_average = ∑ $x_{i}$ / n-1

The angle measurements are in the Sixty Base System, to simplify the calculations let's reduce the measurements

1° = 60 min

1 min = 60 s

Table 1

Angle angle converted (x-x_average) (x-x_average)²

32 32,000 2,51077 6.3040

37 00” 37,000 2,48923 6.1963

32 32,000 2,51077 6.3040

37 10” 37.00277 2.492 6.2101

32 32,000 2,51077 6.3040

37 10” 37.1667 2.6559 7.0538

32 32,000 2,51077 6.3040

36 55” 36.9167 2.4059 5.7884

With this values in degree the calculation is easier

x_average = (32 + 37 + 32 + 37.00277 + 32 + 37.1667 + 32 + 36.9167) / 8

x_average = 276.08617 / 8

x_average = 34.51077

b) We look for the standard deviation

σ = √ ( $x_{i}$ -x_average)² / n-1

σ = √ 50.4646 / (8-1)

σ = 2.69

c) the deviation from the mean

σ = | x - $x_{i}$ | / n

σ = 20.08611 / 8

σ = 2.5107

8 0

3 years ago

A journal bearing has shaft with a diameter of 75 mm. The bushing bore is 75.1 mm and the bushing is 37.5 mm long and supports a

Rainbow [258]

Explanation:

The given data is as follows.

Length (l) = 37.5 mm, Journal diameter (d) = 75 mm

Diameter of bushing (D) = 75.1 mm, Speed (N) = 720 rpm,

Load (F) = 2 kN

Dynamic viscosity of SAE 20 at $60^{o}C$ is 12.2 cp

Clearance = $(\frac{D}{2}) - (\frac{d}{2})$

c = $(\frac{75.1}{2}) - (\frac{75}{2})$

= 0.05 mm

Bearing pressure will be calculated as follows.

Bearing pressure (p) = $\frac{F}{l \times d}$

= $\frac{2 \times 10^{3}}{37.5 \times 75}$

= $0.711 N/mm^{2}$

Now, sommerfield number (S) will be calculated as follows.

S = $(\frac{r}{c})^{2} \times \frac{\mu \times N}{p}$

= $(\frac{37.5}{0.05})^{2} \times \frac{12.2}{10^{9}} \times \frac{720}{60} \times \frac{1}{0.711}$

= 0.006

As, $\frac{l}{d} = \frac{37.5}{75} = \frac{1}{2}$

Now, using Raimondi and John Boyd data the values for other variables will be obtained.

Minimum film thickness variable = $(\frac{h_{o}}{c})$ = 0.03

Therefore, minimum film thickness $(h_{o}) = 0.03 \times 0.05$

= 0.0015 mm

As, $(\frac{P}{P_{max}})$ = 0.126

Hence, maximum lubricant pressure will be calculated as follows.

$P_{max} = \frac{0.711}{0.126}$

= $5.642 N/mm^{2}$

Due to friction, the heat loss rate will be as follows.

$\frac{2 \pi NfFr}{10^{6}}$ kW

According to coefficient of friction variable = $\frac{r}{c} \times f = 0.61$

f = $\frac{c}{r} \times 0.61$

= $\frac{0.05}{37.5} \times 0.61$

= 0.000813

Therefore, heat loss will be calculated as follows.

Heat loss = $\frac{2 \pi \times \frac{720}{60} \times 0.000813 \times 2 \times 10^{3} \times 37.5}{10^{6}}$

= 4.6 Watt

6 0

3 years ago

The pressure of a car can be adjusted using what​

The pressure of a car can be adjusted using what