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3 years ago

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Equation of orbit under central force

1 answer:

SOVA2 [1]3 years ago

8 0

I found this on arxsiv.org: “The central force motion between two bodies about their center of mass can be reduced to an equivalent one body problem in terms of their reduced mass m and their relative radial distance r. ... The potential V (r) from which this force is derived is also a function of r alone, F = −VV, V ≡ V (r).”

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If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

<em>where</em>,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

3 0

3 years ago

a 70 kg desk sits on the floor motionless If gravity is pulling it with an acceleration of 9.8 m/s/s how much force is gravity e

adell [148]

Since force is mass*acceleration,

F = 70kg * 9.8 m/s

3 0

3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

The black hole is___<br> times smaller that the star.<br> I need answers please

mojhsa [17]

Answer:

The answer is 24 (for the first question).

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>

3 0

2 years ago

Which vector correctly indicates the direction of centripetal acceleration experienced by the car?

gogolik [260]

The answer is C.

I hope this helped!

5 0

3 years ago

Read 2 more answers