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2 years ago

Calculate the acceleration of a 2000kg single engine airlane just before take off when the thrust of the engine is 500n?

Engineering

1 answer:

pshichka [43]2 years ago

4 0

Answer:

0.25 m/s²

Explanation:

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An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the g

Harlamova29_29 [7]

Answer:

Gravitational Potential =58.914 KJ

Explanation:

We know that

$Gravitational Potential Energy = mass\times g\times Height$

Given mass = 251 kg

Height= 24 m

g is acceleration due to gravity = $9.78m/s^{2}$

Applying values in the equation we get

$Gravitational Potential Energy=251X9.78X24 Joules$

$Gravitational Potential Energy=58914.72 Joules$

$Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ$

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3 years ago

Whats viruses c liver?

PtichkaEL [24]

Answer:

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2 years ago

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A common process for increasing the moisture content of air is to bubble it through a column of water. The air bubbles are assum

likoan [24]

Answer:

Explanation:

Assumptions is that

1. The flow is an unsteady one

2. Bubbles diameter is constant

3. The bubble velocity is slow

4. There is no homogenous reaction

5. It has a one dimensional flux model along the radial direction

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3 years ago

It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc

ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength $(\sigma_{cd})=630\ MPa = 630*10^6\ Pa$ , Fracture strength

$(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa$

a) The critical length ( $L_c$ ) is given by:

$L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm$

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

$\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456$

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2 years ago

If extension done in a stork position, during backward bending in a standing position of a special test for lumbar spine, produc

Klio2033 [76]

The answer is 7 because I just took the test!

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1 year ago