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Nata [24]
3 years ago
13

What is the difference between the inner and outer planets?

Physics
2 answers:
max2010maxim [7]3 years ago
3 0
The inner planets are the planets before the asteroid belt. They are also closer to the Sun. The outer planets are the ones after the asteroid belt. <span />
Galina-37 [17]3 years ago
3 0
Closer to asteroid belt / sun
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Which characteristics is common in mature rivers river
amm1812
It channels erode wider fed by many tributaries and has more discharge and is less steep
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The distance, in feet, a moving object has traveled after t seconds is given by 2t/(4 + t). find the acceleration of the object
lianna [129]
Hi! Let me help you!
a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0 
a = [(2t)/(4+t)]/t <---- working equation
a = {[2(5)]/9}/5 <---- cancel 5
a = 2/9 ft/s^2 <---- Answer
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3 years ago
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc
lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

E =  \frac{V}{L}

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

J = \frac{E}{p}

Substitute (V/L)  for E in the above equation of current density.

J = \frac{V}{pL} ------(1)

Substitute iR for V in equation (1)

J = \frac{iR}{pL} ------(2)

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

p = \frac{RA}{L}

A = \frac{pL}{R}

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A  and λ is use for (m/L) in the eqn above

\lambda = d\frac{p}{\frac{R}{L} } ------(3)

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³  for d and 1.69 × 10⁸ Ω .m

\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1

c) Using the equation (2) current density for aluminum cable is,

J = \frac{iR}{pL}

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

\lambda = d\frac{p}{\frac{R}{L} }

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0
3 years ago
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So I don’t really get it can you also explain what’s the difference between radiation,conduction, and convection?
Slav-nsk [51]

Answer:

I think its radiation

Explanation:

Conduction is the transfer of heat through solids (A)

Convection is the transfer of heat through liquids or gasses (B)

Radiation is the transfer of heat through em waves (C)

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3 years ago
If Star A and Star B have the same absolute magnitude, but Star A is brighter, what does that tell us?
Hatshy [7]
We can conclude that star A is closer to us than star B.

In fact, the absolute magnitude gives a measure of the brightness of the star, if all the stars are placed at the same distance from Earth. So, it's a measure of the absolute luminosity of the star, indipendently from its distance from us: since the two stars have same absolute magnitude, it means that if they were at same distance from Earth, they would appear with same luminosity. Instead, we see star A brighter than star B, and the only explanation is that star A is closer to Earth than star B (the closer the star A, the brigther it is)
6 0
3 years ago
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