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Sliva [168]
4 years ago
10

A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met

ers above the surface of the Earth. Approximately what percent of its weight does the space shuttle retain?
11%
44%
83%
91%
Physics
1 answer:
faltersainse [42]4 years ago
4 0

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

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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
4 years ago
Read 2 more answers
Ethanethiol (; also called ethyl mercaptan) is commonly added to natural gas to provide the ""rotten egg"" smell of a gas leak.
Tju [1.3M]

Answer:

S = 27500J / 308.15molK

Explanation:

Entropy measures the degree of disorganization of a system. It is measured in the entropy change that is equal to the heat exchanged divided by the temperature at which the process occurs.

S2-S1 = Q / T

S = entropy

Q = heat = 27.5 kJ / mol * 1000J / 1KJ = 27500J / mol

T = temperature = 35 + 273.15 = 308.15K

units = J / molK

S = 27500J / 308.15molK

3 0
3 years ago
Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis
Veronika [31]

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

3 0
3 years ago
Water is an example of ___________, while hydrogen is an example of __________.
-BARSIC- [3]
Are you referring to the fact that water is a compound while hydrogen is an element? If I'm wrong just comment and clarify and I can edit it, I don't even know what kind of unit you're in. :)
3 0
3 years ago
Read 2 more answers
Which of the following tend to react by gaining 1 electron
Andrei [34K]
The correct answer should be C
8 0
3 years ago
Read 2 more answers
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