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larisa86 [58]
3 years ago
14

41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h

as been discovered and has a radius of 7.8 X 10 meters. What is the period of time for Clayton J-21 to orbit Dayli?
Physics
1 answer:
romanna [79]3 years ago
4 0

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

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What is the wave speed of a wave that has a frequency of 250 Hz and a wavelength of 0.35 m?
azamat

Answer:

87.5 m/s

Explanation:

The speed of a wave is given by

v=\lambda f

where

v is the wave speed

\lambda is the wavelength

f is the frequency

In this problem, we have

f=250 Hz is the frequency

\lambda=0.35 m is the wavelength

Substituting into the equation, we find

v=(0.35 m)(250 Hz)=87.5 m/s

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3 years ago
what occurs in a chemical reaction? A. Reactants are formed without chemical bonds being broken B. Reactants are formed from pro
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The answer is D. Products are formed from reactants by the breaking and forming of new bonds. 
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What do extremely massive stars end their life as?
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The most massivest stars end their lives as black holes. <em>(D)</em>

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Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the
IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

\lambda=471\ nm

We need to calculate the distance between the two slits

m\times \lambda=d\sin\theta

d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

d = distance between the two slits

Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

d=40.11\times10^{-6}\ m

d = 40.11\ \mu m

Hence, The distance between the two slits is 40.11 μm.

7 0
3 years ago
The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner
Dahasolnce [82]

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

From lens makers formula

1/f = (n - 1)(1/18 + 1/18) = (n-1)/9

f = 9/(n-1)

Power, P = 1/f ( in m) = (n-1)/0.09

Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09

Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

n = 1.4

7 0
3 years ago
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