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4 years ago

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Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin

g Earth's rotation, how much energy is needed to get the same object into orbit at height h?Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.

1 answer:

Sergio [31]4 years ago

8 0

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

= - GMm/ (R+h)

Potential difference

= GMm/ R - GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.

Extra energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

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The minimum value of magnetic force is 0

Potential V=2.40x $10^{3}$ .

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Since the particle is moving from rest, the potential energy change is 2.40x $10^{3}$ ?

Potential Energy= vq= 2,40x $10^{3}$ x (-1.6 x 10x $10^{-19}$ %).

= -8.84x $10^{-16}$

Now this will be equal to 1/2 mv²

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Wow Fmax = qvb

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=7.88x $10^{-12}$

b) The minimum value of magnetic force

The minimum Value of magnetic force will be

when sinθ=0

The minimum value is 0

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3 years ago

Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold

arsen [322]

Answer:

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Explanation:

In this question we have given

Density of gold, $d=19.3\times 10^3 \frac{kg}{m^3}$

resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

Length of wire, $L= 2.05 km$

Temperature, $T= 20^oC$

We know that relation between volume and density is given as

$Density= \frac{mass}{Volume}$

Therefore, volume occupied by one gram gold is given as,

$V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3$ .........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

$V=\pi \times r^2 \times L$ ......(2)

Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

we got

$V=A \times L$ ...............(4)

on comparing equation 1 and equation 4, we got,

$A \times L=5.181\times 10^{-8} m^3$

$A=\frac{5.181\times 10^{-8} m^3}{2050 m}$

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we know that resistance and resistivity are related by following formula,

$Resistance = resistivity\times \frac{L}{A}$ ................(5)

Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$

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3 years ago