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Sav [38]
4 years ago
5

In 1996, astronomers discovered an icy object beyond Pluto that was given the designation 1996 TL 66. It has a semimajor axis of

84 AU. What is its orbital period according to Kepler’s third law?
Physics
1 answer:
Nat2105 [25]4 years ago
3 0

Answer:

772.65 years.  

Explanation:

Semi major axis = R = 84 AU = 1.257 × 10¹³ m

According to the 3rd law of Kepler, T² =  4 π² R³ / GM

Here R is the semi major axis. G = 6.67 × 10⁻¹¹ SI units.

M is the mass of the  Sun  = 1.98 x 10³⁰ kg

T² =    4 (3.14)³ (1.257 × 10¹³ )³÷ (6.67 × 10⁻¹¹)(1.98 x 10³⁰)

  ⇒  Time period  =  T = 2.44 x 10¹⁰ seconds

1 s = 3.17 × 10⁻⁸ years

Converting the seconds to years, T = 772.65 years.

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A violinist is tuning her violin to 440Hz. She plays the note while listening to an electronically generated tone and hears 3Hz,
BartSMP [9]

Answer:

X=438Hz

Explanation:

From the question we are told that:

Beat Frequency Fb=440Hz

Actual Frequency F_a=3Hz

Generally the equation for Frequency Heard X is mathematically given by

X=beat\ frequency\ +\ actual\ frequency

X=F_a+Fb

X= 440Hz+3Hz

X=438Hz

5 0
3 years ago
The coordinate of a particle in meters is given by x(t) = 16t − 3.0t3, where the time t is in
Advocard [28]

Answer:

t = -4/3 s, t = 4/3 s

Explanation:

I am ASSUMING that you want to know the time(s) the particle is at rest.

and that your position equation is x(t) = 16t − 3.0t³

IF those assumptions are true

THEN

velocity is the derivative of position

v(t) = 16 - 9t²

the particle will be at rest when velocity is zero

0 = 16 - 9t²

9t² = 16

t² = 16/9

t = ± 4/3

7 0
3 years ago
A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per m
Natali [406]

(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.

(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.

<h3>Electric field at a point on the Gaussian surface</h3>

The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;

E = λ/2πε₀r

where;

  • λ is linear charge density
  • ε₀ is permitivity of free space
  • r is the position of the charge
<h3>At a distance of 5.5 m</h3>

E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C

<h3>At a distance of 2.5 m</h3>

E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C

Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.

Learn more about electric field here: brainly.com/question/14372859

4 0
2 years ago
Estimate the peak wavelength for radiation from ice at 273 k.
Andrews [41]
<h2>Answer: 10615 nm</h2>

Explanation:

This problem can be solved by the Wien's displacement law, which relates the wavelength  \lambda_{p} where the intensity of the radiation is maximum (also called peak wavelength) with the temperature T of the black body.

In other words:

<em>There is an inverse relationship between the wavelength at which the emission peak of a blackbody occurs and its temperature.</em>

Being this expresed as:

\lambda_{p}.T=C    (1)

Where:

T is in Kelvin (K)

\lambda_{p} is the <u>wavelength of the emission peak</u> in meters (m).

C is the <u>Wien constant</u>, whose value is 2.898(10)^{-3}m.K

From this we can deduce that the higher the black body temperature, the shorter the maximum wavelength of emission will be.

Now, let's apply equation (1), finding \lambda_{p}:

\lambda_{p}=\frac{C}{T}   (2)

\lambda_{p}=\frac{2.898(10)^{-3}m.K}{273K}  

Finally:

\lambda_{p}=10615(10)^{-9}m=10615nm  This is the peak wavelength for radiation from ice at 273 K, and corresponds to the<u> infrared.</u>

8 0
3 years ago
What is the frequency of an X-ray with wavelength 0.13 nm ? Assume that the wave travels in free space. Express your answer to t
dem82 [27]

Answer:

Frequency, f=2.30\times 10^{18}\ Hz

Explanation:

Given that,

The wavelength of the x-rays, \lambda=0.13\ nm=0.13\times 10^{-9}\ m

We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

c=f\lambda

f is the frequency

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.13\times 10^{-9}}\\\\f=2.30\times 10^{18}\ Hz

So, the frequency of an x-ray is 2.30\times 10^{18}\ Hz. Hence, this is the required solution.

5 0
3 years ago
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