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3 years ago

Which of the following can easily reverse motion and are better at varying speeds than electrical motors?

Engineering

2 answers:

yan [13]3 years ago

6 0

Answer:

fluid power system

Explanation:

satela [25.4K]3 years ago

4 0

Fluid power system

Example

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Is a diesel truck less expensive to drive than a gas truck?

MArishka [77]

Answer:

Typically, diesel trucks cost more than those with gas engines, especially when you're first buying them, as diesel is usually featured as an add-on for gas-powered cars. Diesel add-ons can cost over $5,000 for midsize trucks and around $10,000 for heavy-duty trucks.

Explanation:

Make me brain pls

3 0

2 years ago

It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d

denis23 [38]

The exit temperature is 586.18K and compressor input power is 14973.53kW

Data;

Mass = 50kg/s
T = 288.2K
P1 = 1atm
P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\$

Let's substitute the values into the formula

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K$

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

$P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW$

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

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6 0

3 years ago

A platform and a weight totaling 900 N is held in position by a vertical water jet of 5 cm diameter. Determine the velocity of t

BlackZzzverrR [31]

Answer:

v = 21.409 m/s

Explanation:

Given data:

Total weight of the platform and weight together, W = 900 N

Diameter of the water jet = 5 cm = 0.05 m

Now,

the force exerted by the water jet is balancing the platform

also,

Force exerted by the water jet = ρAv²

where,

ρ is the density of the water = 1000 kg/m³

A is the area of the outlet of the jet = $\frac{\pi}{4}d^2=\frac{\pi}{4}(0.05)^2$

A = 0.00196 m²

v is the velocity of the jet

thus,

W = ρAv²

900 = 1000 × 0.00196 × v²

v = 21.409 m/s

Hence, the velocity of the jet is 21.409 m/s

7 0

3 years ago

On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch

rewona [7]

Answer:

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

Explanation:

Distance of earth from sun = $R_{2} = 147.4 \times 106 Km$

Spacecraft perihelion = $R_{2} = 120\times106Km$

gravitational parameters are now given as

$\mu_{sun} = 132.7\times 10^{9}$

$\mu_{earth} = 398600$

radius of earth = 6378 Km

Heliocentric spacecraft velocity at earth sphere of influence =

$V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }$

$V_{D}^{v} =28.43\frac{km}{s}$

Heliocentric velocity of earth = $v_{earth} = 30.06\frac{km}{sec}$

$V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}$

assume

$r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km$

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

$v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}$

geometric space craft velocity in its circular parking orbit

$v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } } = 7.784 \frac{km}{s}$

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

7 0

4 years ago

Cutting and forming operations performed on relatively thin sheets of metal. a)- True b)-False

belka [17]

Answer: True

Explanation: There are different actions that are performed on the metal sheet . Cutting operation is the operation of splitting,slitting or parting of the metal sheet on the by using different kind of blades . Formation operation are basically for the exerting pressure on the sheet to give it desired shape such as bending, squeezing etc. Therefore the given statement is true.

6 0

3 years ago