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3 years ago

Kepler's 3rd law: harmonic law

Physics

1 answer:

Amanda [17]3 years ago

7 0

<span>orbital velocities to their mean distances from the Sun.</span>

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A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

A 1,000 kg ball traveling at 5 m/s would have kinetic energy

MatroZZZ [7]

Answer:

12500 J = 12.5 kJ

Explanation:

Kinetic energy is the energy possessed by a moving object solely due to its motion.

You can get the K.E. of an object using the equation,

K.E. = (1/2)mv²

So you get, for ball

K.E. = (1/2)×1000×5² = 12500 J = 12.5 kJ

6 0

3 years ago

The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per

Natasha_Volkova [10]

Answer:

0.2631 N/C

Explanation:

Given that:

The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m

The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m

The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons

Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)

= 0.0006

The number of electron flow per second is calculated as follows:

$I = \frac{q}{t}$

$I = \frac{Ne}{t}$

$I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}$

$I = 0.96 \ A$

The magnitude of the electric field is:

E = $\frac{I}{ \mu n eA}$

E = $\frac{I}{ \mu n e(\pi r^2)}$

E = $\frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}$

E = 0.2631 N/C

8 0

3 years ago

A water tank holds water to a depth of 80 cm.

Vlad1618 [11]

Answer:

8000 pa

Explanation:

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8 0

3 years ago

How many turns are in the secondary coil of a step up transformer that increases voltage from 30 V to 150 V and has seven turns

Mademuasel [1]

350 + 507 = 3657 that would be your answer

7 0

3 years ago