Why does a foil airplane fly farther than a paper one?

The ________________ attraction between the Earth and the moon is ______________ on the side of the Earth that happens to be ___

Karolina [17]

Answer:

Gravitational; strongest; facing; closer; near side; toward.

Explanation:

The gravitational attraction between the Earth and the moon is strongest on the side of the Earth that happens to be facing the moon, simply because it is closer. This attraction causes the water on this “near side” of Earth to be pulled toward the moon. These forces of attraction and inertia tends to keep the water in place and consequently, leads to a bulge of water on the near side with respect to the moon.

Also, you should note that what is responsible for the moon being in orbit around the Earth is the gravitational force of attraction between the two planetary bodies (Earth and Moon).

7 0

3 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

5 0

3 years ago

A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic

Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0

3 years ago

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 $m^3/kg$

Specific volume of saturated liquid=v_f=0.00102 $m^3/kg$

Now the relation of total specific volume for a specific dryness value is given as

$v=v_f+x(v_g-v_f)$

Substituting the values give

$v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg$

Now the volume of vessel is given as

$v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3$

So the volume of vessel is 4.7680 $m^3$ .

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ$

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

$v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg$

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ$

So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0

3 years ago

Can someone answer plz!! It’s 24 points

fgiga [73]

Explanation:

750 microvolt is your answer

please mark as brilliant

3 0

3 years ago

Read 2 more answers