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Nostrana [21]
3 years ago
9

How long does it take the statue to reach the ground ?

Physics
1 answer:
Marrrta [24]3 years ago
8 0
The Statue of Liberty?
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What is the average time in days between moon phases?.
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Answer:

28

Explanation:

There is a 28 day cycle of the moon phase as it rotates on earths axis.

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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

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3 years ago
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What is the formula for mechanical advantage?
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Answer:

A. MA=force output/force input

Explanation:

mechanical advantage is the ratio of the load to the effort

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3 years ago
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A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5
podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity (\omega_o) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity  (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\

b)

\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)}  =112.4\ rev

θ = 323 rad

4 0
2 years ago
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