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3 years ago

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During the collision, is the magnitude of the force of asteroid A on asteroid B greater than, less than, or equal to the magnitu

de of the force of asteroid B on asteroid A?

2 answers:

Natasha2012 [34]3 years ago

5 0

Answer: The magnitude of the fore of asteroid A on asteroid B is equal to the magnitude of the force of asteroid B on asteroid A.

Explanation:

The force of one asteroid on the other, is explained by the Universal Law of gravitation, that can be written as follows:

Fg = G* mA*mB / (rAB)²

Even though the force is a vector, and is directed towards one of the masses, along the line that joins their centers, it can be seen the magnitude is the same.

This can be explained also stating that these forces form a pair of action-reaction forces, which, according with Newton’s 3rd Law, must be equal and opposite each other.

goblinko [34]3 years ago

3 0

Answer:

B

Explanation:

greater than, less than, or equal to the magnitude of the force of asteroid!

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Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

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3 years ago

alexandr402 [8]

Answer:

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Explanation:

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3 years ago

Which of the following procedures best applies to assessing the embodied energy of a building?

galina1969 [7]

analyzing building materials???????????? but i can try i think it is analyzing the materials of the building

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2 years ago

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

4 0

3 years ago

Steam enters an adiabatic turbine operating at steady state at 2MPa, 400 C with velocity of 50 m/s and exit at 15 KPasquality of

goldenfox [79]

Answer:

P=4833.45 KW

Explanation:

At inlet:

Properties of steam at 2 MPa and 400°C

$h_1=3248.22\frac{KJ}{Kg}$

$V_1=50 m/s$

At exit:

Properties of steam at 15 KPa

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

So at 15 KPa

$h_f= 33.61\frac{KJ}{Kg} ,h_g=2515.2\frac{KJ}{Kg}$

$V_2=180m/s$

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=33.61+0.9(2515.2-33.61)\frac{KJ}{Kg}$ $h_2=2267.04\frac{KJ}{Kg}$

Now from first law of thermodynamics for open system

$h_1+\dfrac{V_1^2}{2}+Z_1g+Q=h_2+\dfrac{V_2^2}{2}+Z_2g+w$

$Z_1=47 m,Z_2=0$

Here given that turbine is adiabatic so Q=0

Now put the all values

$3248.22+\dfrac{50^2}{2000}+47\times 9.81\times 10^{-3}+0=2267.04+\dfrac{180^2}{2000}+W$

$W=966.69\frac{KJ}{Kg}$

$So power developed P=m\times W$

$P=5\times966.69$

P=4833.45 KW

5 0

3 years ago